CUET UG Applied Mathematics Booster Test 2 - Integration by Substitution -

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CUET UG Applied Mathematics Booster Test 2 - Integration by Substitution

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Applied Mathematics Booster Test 2 - Integration by Substitution

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QUESTION 1 OF 20

Which specific variable substitution mathematically converts the integral
\(\int \frac{1}{x(1+log⁡x)^{2}} dx\)
into a standard power rule evaluation form?

Answer Review

1 Which specific variable substitution mathematically converts the integral
\(\int \frac{1}{x(1+log⁡x)^{2}} dx\)
into a standard power rule evaluation form?

A Put \(x=t\)

B Put \(\frac{1}{x}=t\)

C Put \(1+log⁡x=t\)

D Put \(log⁡x=t^{2}\)

💡 Short Explanation

Identify the inner expression. Differentiate \(1+log⁡x\). The integral becomes a simple power integral.

✅ Correct Answer Explanation

Let \(t=1+log⁡x.\) Then, \(dt=\frac{dx}{x}.\) The integral transforms into \(\int \frac{1}{t^{2}} dt=\int t^{-2} dt,\) which is a standard power-rule integral. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Does not simplify the composite logarithmic expression.
Option B → Produces a more complicated differential.
Option D → Squaring the logarithm unnecessarily complicates the substitution.

🎯 Strategy

Used

Substitution

Application:

Choose the entire inner function whose derivative is present in the integrand.

Final Logic:

Since
\(d(1+log⁡x)=\frac{dx}{x},\)
the substitution directly converts the integral into a power form.

🧠 Memory Trick

"Inside Function + Derivative = Best Substitution."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Integration by Substitution (Composite Logarithmic Functions)

2 Match the indefinite integral in List I with its correctly evaluated result utilizing standard substitution in List II.

List I	List II
1. \(\int \frac{2x}{x^{2}+1} dx\)	a. \(\sqrt{x^{2}-1}+C\)
2. \(\int e^{2x} dx\)	b. \(log⁡∣x^{2}+1∣+C\)
3. \(\int \frac{x}{\sqrt{x^{2}-1}} dx\)	c. \(\frac{1}{2}log⁡∣2x+3∣+C\)
4. \(\int \frac{1}{2x+3} dx\)	d. \(\frac{e^{2x}}{2}+C\)

A 1–a, 2–b, 3–c, 4–d

B 1–c, 2–a, 3–d, 4–b

C 1–b, 2–d, 3–a, 4–c

D 1–d, 2–c, 3–b, 4–a

💡 Short Explanation

Match each integral with its standard substitution result. Recognize logarithmic, exponential and radical forms. Use standard integration formulas.

✅ Correct Answer Explanation

The correct matches are: 1 → b \(\int \frac{2x}{x^{2}+1}dx=log⁡∣x^{2}+1∣+C\) 2 → d \(\int e^{2x}dx=\frac{e^{2x}}{2}+C\) 3 → a \(\int \frac{x}{\sqrt{x^{2}-1}}dx=\sqrt{x^{2}-1}+C\) 4 → c \(\int \frac{1}{2x+3}dx=\frac{1}{2}log⁡∣2x+3∣+C\) Thus, \(1-b, 2-d, 3-a, 4-c\) Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrectly matches every integral.
Option B → Logarithmic and exponential results are mismatched.
Option D → Composite substitutions are incorrectly paired.

🎯 Strategy

Used

Option Grouping

Application:

Recall the standard substitution result for each integral and systematically match them.

Final Logic:

Each integral has a unique standard antiderivative leading to Option C.

🧠 Memory Trick

"Derivative Over Function → Log, Linear Exponent → Divide, Radical → Raise Power."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Standard Integrals Using Substitution

3 When accurately evaluating the definite integral
\(\int_{0}^{2}\,t\sqrt{t^{2}+1} dt\)
by executing the substitution
\(t^{2}+1=u,\)
which of the following transformational steps are correct?
1. The differential is translated via \(2t dt=du\).
2. The new numerical lower limit correctly becomes \(1\).
3. The new numerical upper limit correctly becomes \(5\).
4. The transformed integral becomes
\(\frac{1}{2}\int u^{1/2} du.\)

A 1, 2

B 2, 3, 4

C 1, 2, 3

D 1, 2, 3, 4

💡 Short Explanation

Differentiate the substitution. Change both limits. Rewrite the integral completely in terms of \(u\).

✅ Correct Answer Explanation

Let \(u=t^{2}+1.\) Then, \(du=2t dt.\) Hence, \(t dt=\frac{du}{2}.\) Limits become: When \(t=0\), \(u=1.\) When \(t=2\), \(u=5.\) Therefore, \(\int_{0}^{2}\,t\sqrt{t^{2}+1} dt=\frac{1}{2}\int_{1}^{5}\,u^{1/2} du.\) Thus, all four statements are correct, making Option D the correct answer.

❓ Why Other Options Are Incorrect

Option A → Omits Statements 3 and 4.
Option B → Omits Statement 1.
Option C → Omits the transformed integral expression.

🎯 Strategy

Used

Option Grouping

Application:

Verify each statement individually before selecting the combination containing all correct statements.

Final Logic:

All four statements follow directly from the substitution process.

🧠 Memory Trick

"Substitute → Differentiate → Change Limits → Rewrite Integral."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Definite Integrals by Substitution (Multiple Correct Questions)

4 Which of the following statements is INCORRECT regarding fractional power and standard variable substitution rules?

A The mathematical conversion to \(dt\) logically accounts for the derivative of the inner function.

B Fractional numerical powers can never be integrated without relying on complex trigonometric substitutions.

C Formal substitution effectively transforms radical expressions into computationally manageable polynomial forms.

D For an integrand with \({\left(\right)}^{m/n}\), substituting \(f(x)=t^{n}\)eliminates the fractional algebraic exponent.

💡 Short Explanation

Fractional powers are often simplified using algebraic substitution. Trigonometric substitution is needed only for specific forms. The statement uses the absolute word "never," making it incorrect.

✅ Correct Answer Explanation

The statement "Fractional numerical powers can never be integrated without relying on complex trigonometric substitutions." is incorrect because many fractional power integrals are solved using simple algebraic substitution, such as \(f(x)=t^{n},\) which converts fractional exponents into whole-number powers. Trigonometric substitution is required only for specific radical expressions such as \(\sqrt{a^{2}-x^{2}},\sqrt{x^{2}+a^{2}},\sqrt{x^{2}-a^{2}},\) not for all fractional powers. Therefore, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A → Correct. During substitution, \(dt\) is obtained by differentiating the substituted expression.
Option C → Correct. Substitution often converts radical expressions into polynomial forms.
Option D → Correct. Letting \(f(x)=t^{n}\)removes the fractional exponent and simplifies integration.

🎯 Strategy

Used

Extreme Word Filter

Application:

Words such as "never," "always," "only," and "all" often indicate incorrect statements in conceptual questions. Verify such statements carefully.

Final Logic:

The word "never" makes the statement universally false because many fractional power integrals are solved without trigonometric substitution.

🧠 Memory Trick

"Extreme words? Check twice!"

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Standard Methods of Integration by Substitution

5 Evaluate the continuous algebraic substitution mixture:
\(\int (x+2)\sqrt{x-1} dx\)
by putting
\(x-1=t^{2}.\)

A \(\frac{{\left(x-1\right)}^{3/2}}{3}+C\)

B \(\frac{{\left(x-1\right)}^{5/2}}{5}+C\)

C \(\frac{2}{5}(x-1)^{5/2}+2(x-1)^{3/2}+C\)

D \(\frac{2}{5}(x-1)^{5/2}+(x-1)^{3/2}+C\)

💡 Short Explanation

Substitute \(x-1=t^{2}\). Rewrite the entire integrand in terms of \(t\). Integrate the resulting polynomial and substitute back.

✅ Correct Answer Explanation

Let \(x-1=t^{2}.\) Then, \(x=t^{2}+1,\) and \(dx=2t dt.\) Also, \(x+2=t^{2}+3,\sqrt{x-1}=t.\) Hence, \(\int (x+2)\sqrt{x-1} dx=\int (t^{2}+3)t(2t) dt=2\int (t^{4}+3t^{2}) dt.\) Integrating, \(2\left(\frac{t^{5}}{5},\ t^{3}\right)+C=\frac{2}{5}t^{5}+2t^{3}+C.\) Replacing \(t=\sqrt{x-1},\) gives \(\frac{2}{5}(x-1)^{5/2}+2(x-1)^{3/2}+C.\) Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Omits the higher-degree term resulting from the substitution.
Option B → Ignores the contribution of the \(2(x-1)^{3/2}\)term.
Option D → The coefficient of \({\left(x-1\right)}^{3/2}\)should be 2, not 1.

🎯 Strategy

Used

Substitution

Application:

Convert the radical into a polynomial by substituting \(x-1=t^{2}\), then integrate term by term.

Final Logic:

After substitution, the integral becomes a polynomial integral whose evaluation gives Option C.

🧠 Memory Trick

"Radical → Square Variable → Polynomial Integral."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Integration by Algebraic Substitution (Radical Functions)

6 Accurately evaluate the integral reliant on quadratic expression constraints:
\(\int \frac{1}{\sqrt{9+4x^{2}}} dx\)
using standard root formulas.

A \(\frac{1}{2}{sin⁡}^{-1} \left(\frac{2x}{3}\right)+C\)

B \(\frac{1}{3}log⁡∣x+\sqrt{9+4x^{2}}∣+C\)

C \(\frac{1}{2}log⁡∣2x+\sqrt{9+4x^{2}}∣+C\)

D \(log⁡∣2x+\sqrt{9+4x^{2}}∣+C\)

💡 Short Explanation

Rewrite the integrand in the standard form \(\int \frac{dx}{\sqrt{a^{2}+b^{2}x^{2}}}\). Apply the logarithmic standard integral. Simplify the result.

✅ Correct Answer Explanation

Given, \(I=\int \frac{1}{\sqrt{9+4x^{2}}} dx.\) Using the standard result, \(\int \frac{dx}{\sqrt{a^{2}+b^{2}x^{2}}}=\frac{1}{b}log⁡∣bx+\sqrt{a^{2}+b^{2}x^{2}}∣+C,\) where \(a=3\) and \(b=2\), \(I=\frac{1}{2}log⁡∣2x+\sqrt{9+4x^{2}}∣+C.\) Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → This inverse sine formula applies to \(\sqrt{a^{2}-x^{2}}\), not \(\sqrt{a^{2}+x^{2}}\).
Option B → The coefficient should be \(\frac{1}{2}\), not \(\frac{1}{3}\).
Option D → Omits the required factor \(\frac{1}{2}\).

🎯 Strategy

Used

Substitution

Application:

Recognize the standard radical form and apply the corresponding logarithmic integral formula.

Final Logic:

The standard formula directly yields Option C.

🧠 Memory Trick

"\(\sqrt{a^{2}+x^{2}}\) → Log appears."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Standard Integrals Involving \(\sqrt{a^{2}+x^{2}}\)

7 Evaluate the following advanced logarithmic and exponential nested integral:
\(\int 3^{\left(3^{x}\right)} 3^{x} dx.\)

A \(\frac{3^{3^{x}}}{\log\,3}+C\)

B \(\frac{3^{x}}{{\left(log⁡3\right)}^{2}}+C\)

C \(\frac{3^{3^{x}}}{{\left(log⁡3\right)}^{2}}+C\)

D \(3^{x^{2}}log⁡3+C\)

💡 Short Explanation

Let \(t=3^{x}\). Then \(dt=3^{x}log⁡3 dx\). Integrate the exponential in terms of \(t\).

✅ Correct Answer Explanation

Let \(t=3^{x}.\) Then, \(dt=3^{x}log⁡3 dx\Rightarrow 3^{x} dx=\frac{dt}{\log\,3}.\) Hence, \(\int 3^{3^{x}}3^{x} dx=\frac{1}{\log\,3}\int 3^{t} dt.\) Since \(\int 3^{t} dt=\frac{3^{t}}{\log\,3}+C,\) we obtain \(\frac{3^{3^{x}}}{{\left(log⁡3\right)}^{2}}+C.\) Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → One factor of \(\log\,3\) is missing.
Option B → Uses the wrong exponential expression.
Option D → Does not result from substitution.

🎯 Strategy

Used

Substitution

Application:

Substitute the inner exponential function first, then apply the exponential integration formula.

Final Logic:

Two factors of \(\log\,3\) appear—one from substitution and one from integrating \(3^{t}\).

🧠 Memory Trick

"Nested exponent → Two logarithms in the denominator."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Integration by Substitution (Composite Exponential Functions)

8 Evaluate the specific continuous fractional exponential integral:
\(\int \frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}} dx.\)

A \(log⁡∣e^{2x}-e^{-2x}∣+C\)

B \(\frac{1}{2}log⁡∣e^{2x}+e^{-2x}∣+C\)

C \(e^{2x}+e^{-2x}+C\)

D \(\frac{1}{2}log⁡∣e^{2x}-e^{-2x}∣+C\)

💡 Short Explanation

Let \(t=e^{2x}-e^{-2x}\). Its derivative appears in the numerator. Apply the logarithmic integral formula.

✅ Correct Answer Explanation

Let \(t=e^{2x}-e^{-2x}.\) Then, \(dt=2(e^{2x}+e^{-2x}) dx.\) Therefore, \((e^{2x}+e^{-2x}) dx=\frac{dt}{2}.\) The integral becomes \(\frac{1}{2}\int \frac{dt}{t}=\frac{1}{2}log⁡∣t∣+C.\) Substituting back, \(\frac{1}{2}log⁡∣e^{2x}-e^{-2x}∣+C.\) Hence, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Missing the factor \(\frac{1}{2}\).
Option B → Uses the wrong substitution.
Option C → Is not an antiderivative of the given function.

🎯 Strategy

Used

Substitution

Application:

Recognize that the numerator is proportional to the derivative of the denominator.

Final Logic:

Since
\(dt=2(e^{2x}+e^{-2x})dx,\)
the answer is \(\frac{1}{2}log⁡∣t∣+C\).

🧠 Memory Trick

"Derivative over Function → Log, then divide by the extra constant."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Integration by Substitution (Logarithmic Forms)

9 Evaluate the composite probability-bound algebraic integral
\(\int 2x\sqrt{x^{2}+1} dx\)
using standard substitution.

A \(\sqrt{x^{2}+1}+C\)

B \(\frac{1}{3}(x^{2}+1)^{3}+C\)

C \(\frac{2}{3}(x^{2}+1)^{3/2}+C\)

D \(\frac{3}{2}(x^{2}+1)^{2/3}+C\)

💡 Short Explanation

Let \(t=x^{2}+1\). Then \(dt=2x dx\). Integrate \(t^{1/2}\).

✅ Correct Answer Explanation

Given, \(\int 2x\sqrt{x^{2}+1} dx.\) Let \(t=x^{2}+1.\) Then, \(dt=2x dx.\) Hence, \(\int 2x\sqrt{x^{2}+1} dx=\int t^{1/2} dt=\frac{2}{3}t^{3/2}+C.\) Substituting back, \(\frac{2}{3}(x^{2}+1)^{3/2}+C.\) Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → This differentiates to \(\frac{x}{\sqrt{x^{2}+1}}\), not the given integrand.
Option B → Incorrect power after integration.
Option D → Uses an incorrect exponent and coefficient.

🎯 Strategy

Used

Substitution

Application:

Recognize that the derivative of the inner quadratic appears in the numerator.

Final Logic:

Since \(dt=2x dx\), the integral reduces to \(\int t^{1/2}dt\).

🧠 Memory Trick

"Derivative of Inside + Root = Direct Substitution."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Integration by Substitution (Composite Algebraic Functions)

10 Evaluate the nested composite function mapped integral
\(\int xe^{x^{2}} dx\)
using the substitution
\(x^{2}=t.\)

A \(x^{2}e^{x}+C\)

B \(\frac{1}{2}e^{x^{2}}+C\)

C \(2xe^{x^{2}}+C\)

D \(e^{x^{2}}+C\)

💡 Short Explanation

Let \(t=x^{2}\). Then \(dt=2x dx\). Integrate the exponential function.

✅ Correct Answer Explanation

Let \(t=x^{2}.\) Then, \(dt=2x dx\Rightarrow x dx=\frac{dt}{2}.\) Hence, \(\int xe^{x^{2}} dx=\frac{1}{2}\int e^{t} dt=\frac{1}{2}e^{t}+C.\) Replacing \(t=x^{2}\), \(\frac{1}{2}e^{x^{2}}+C.\) Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Does not differentiate to the given integrand.
Option C → Multiplies instead of dividing by the derivative.
Option D → Omits the factor \(\frac{1}{2}\).

🎯 Strategy

Used

Substitution

Application:

Substitute the inner quadratic expression and divide by its derivative.

Final Logic:

Since \(dt=2x dx\), the integral becomes \(\frac{1}{2}\int e^{t} dt\).

🧠 Memory Trick

"Linear derivative with exponential ⇒ Divide by 2."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Integration by Substitution (Exponential Functions)

11 Determine the accurately evaluated area calculation logic for the radical substitution integral
\(\int \frac{x}{\sqrt{x-1}} dx\)
by definitively putting
\(x-1=t^{2}.\)

A \(2(x-1)^{3/2}+C\)

B \(\frac{2}{3}(x-1)+2x+C\)

C \(\frac{1}{3}(x-1)^{3/2}+\sqrt{x-1}+C\)

D \(\frac{2}{3}(x-1)^{3/2}+2\sqrt{x-1}+C\)

💡 Short Explanation

Substitute \(x-1=t^{2}\). Express \(x\), \(dx\), and the radical in terms of \(t\). Integrate the resulting polynomial.

✅ Correct Answer Explanation

Let \(x-1=t^{2}.\) Then, \(x=t^{2}+1,dx=2t dt,\sqrt{x-1}=t.\) Therefore, \(\int \frac{x}{\sqrt{x-1}}dx=\int \frac{t^{2}+1}{t}(2t) dt=2\int (t^{2}+1) dt.\) Integrating, \(2\left(\frac{t^{3}}{3},\ t\right)+C=\frac{2}{3}t^{3}+2t+C.\) Replacing \(t=\sqrt{x-1}\), \(\frac{2}{3}(x-1)^{3/2}+2\sqrt{x-1}+C.\) Hence, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Omits the \(2\sqrt{x-1}\)term.
Option B → Does not result from the substitution.
Option C → Incorrect coefficients for both terms.

🎯 Strategy

Used

Substitution

Application:

Convert the radical into a polynomial using \(x-1=t^{2}\).

Final Logic:

After substitution, the integral becomes a simple polynomial integral.

🧠 Memory Trick

"Root Expression → Square Variable → Polynomial Integral."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Integration by Radical Substitution

12 Calculate the evaluated result for the fractional power substitution integral
\(\int x^{2}\sqrt{x^{3}+1} dx.\)

A \(\frac{2}{3}(x^{3}+1)+C\)

B \(\frac{1}{3}(x^{3}+1)^{3/2}+C\)

C \(\frac{2}{9}(x^{3}+1)^{3/2}+C\)

D \(\frac{2}{9}(x^{3}+1)^{1/2}+C\)

💡 Short Explanation

Let \(t=x^{3}+1\). Then \(dt=3x^{2}dx\). Integrate the resulting power function.

✅ Correct Answer Explanation

Let \(t=x^{3}+1.\) Then, \(dt=3x^{2}dx\Rightarrow x^{2}dx=\frac{dt}{3}.\) Hence, \(\int x^{2}\sqrt{x^{3}+1} dx=\frac{1}{3}\int t^{1/2}dt.\) Integrating, \(\frac{1}{3}⋅\frac{2}{3}t^{3/2}=\frac{2}{9}t^{3/2}+C.\) Substituting back, \(\frac{2}{9}(x^{3}+1)^{3/2}+C.\) Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect power after integration.
Option B → Missing the factor \(\frac{2}{3}\).
Option D → Power remains unchanged after integration.

🎯 Strategy

Used

Substitution

Application:

Recognize that the derivative of the inner cubic is present in the integrand.

Final Logic:

The substitution converts the integral into a standard power-rule integral.

🧠 Memory Trick

"Derivative Present + Root ⇒ Substitute."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Integration by Substitution (Composite Algebraic Functions)

13 Apply the rationalization integration method on linear roots and evaluate
\(\int \frac{1}{\sqrt{x+4}-\sqrt{x-3}} dx.\)

A \(\frac{2}{3}(x+4)^{3/2}-\frac{2}{3}(x-3)^{3/2}+C\)

B \(\frac{1}{7}log⁡∣x+4∣+C\)

C \(7\sqrt{x+4}+C\)

D \(\frac{2}{21}(x+4)^{3/2}+\frac{2}{21}(x-3)^{3/2}+C\)

💡 Short Explanation

Rationalize the denominator using the conjugate. Simplify the resulting expression. Integrate each radical term separately.

✅ Correct Answer Explanation

Multiply the numerator and denominator by the conjugate: \(\frac{1}{\sqrt{x+4}-\sqrt{x-3}}\times \frac{\sqrt{x+4}+\sqrt{x-3}}{\sqrt{x+4}+\sqrt{x-3}}.\) The denominator becomes \((x+4)-(x-3)=7.\) Hence, \(\int \frac{1}{\sqrt{x+4}-\sqrt{x-3}}dx=\frac{1}{7}\int \left(\sqrt{x+4},\ \sqrt{x-3}\right)dx.\) Using \(\int \sqrt{x+a} dx=\frac{2}{3}(x+a)^{3/2},\) we obtain \(\frac{2}{21}(x+4)^{3/2}+\frac{2}{21}(x-3)^{3/2}+C.\) Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Uses subtraction instead of addition after rationalization.
Option B → The integral does not simplify to a logarithmic form.
Option C → Ignores the contribution from \(\sqrt{x-3}\)and the integration process.

🎯 Strategy

Used

Substitution

Application:

First rationalize the denominator to simplify the integrand, then apply the standard radical integration formula.

Final Logic:

Rationalization converts the integral into the sum of two standard radical integrals.

🧠 Memory Trick

"Root Difference → Use the Conjugate."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Integration by Rationalization

14 Evaluate the definite limits transformation:
\(\int_{0}^{3}\,x\sqrt{x+4} dx.\)
By setting
\(x+4=t,\)
what exact numerical integral must now be solved in terms of \(t\)?

A \(\int_{0}^{3}\,(t-4)\sqrt{t} dt\)

B \(\int_{1}^{4}\,t\sqrt{t} dt\)

C \(\int_{4}^{7}\,(t-4)\sqrt{t} dt\)

D \(\int_{4}^{7}\,t\sqrt{t-4} dt\)

💡 Short Explanation

Substitute \(t=x+4\). Convert both limits. Rewrite the integrand completely in terms of \(t\).

✅ Correct Answer Explanation

Let \(t=x+4.\) Then, \(x=t-4,dx=dt.\) The limits become When \(x=0, t=4\). When \(x=3, t=7\). Thus, \(\int_{4}^{7}\,(t-4)\sqrt{t} dt.\) Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Uses the original limits instead of the transformed limits.
Option B → Both the limits and integrand are incorrect.
Option D → The substitution has not been correctly applied to the radical.

🎯 Strategy

Used

Substitution

Application:

Change both the variable and the limits simultaneously for a definite integral.

Final Logic:

Substituting \(t=x+4\) changes the interval from \(\left[0,\ 3\right]\)to \(\left[4,\ 7\right]\).

🧠 Memory Trick

"New Variable → New Limits."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Definite Integrals by Substitution

15 By identifying the exact derivative of the denominator in the numerator, structurally evaluate the integral
\(\int \frac{2x-3}{x^{2}-3x-18} dx.\)

A \(\frac{1}{2}log⁡∣x^{2}-3x-18∣+C\)

B \((x^{2}-3x-18)^{2}+C\)

C \(2xlog⁡∣x^{2}-3x-18∣+C\)

D \(log⁡∣x^{2}-3x-18∣+C\)

💡 Short Explanation

Let \(f(x)=x^{2}-3x-18\). The numerator equals \(f^{'}(x)\). Apply the standard logarithmic integration formula.

✅ Correct Answer Explanation

Let \(f(x)=x^{2}-3x-18.\) Then, \(f^{'}(x)=2x-3,\) which exactly matches the numerator. Using \(\int \frac{f^{'}(x)}{f(x)}dx=log⁡∣f(x)∣+C,\) we get \(log⁡∣x^{2}-3x-18∣+C.\) Hence, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → The factor \(\frac{1}{2}\)is unnecessary because the numerator already equals the derivative of the denominator.
Option B → Differentiating this expression does not reproduce the given integrand.
Option C → Multiplication by \(2x\) is not part of the logarithmic integration formula.

🎯 Strategy

Used

Substitution

Application:

Recognize the standard form \(\frac{f^{'}(x)}{f(x)}\).

Final Logic:

Since the numerator is exactly the derivative of the denominator, the answer is the natural logarithm of the denominator.

🧠 Memory Trick

"Derivative over Function → Natural Log."

📚 Bottom of Form

16 When evaluating the indefinite integral
\(\int \frac{x}{\sqrt{x-1}} dx\)
utilizing the substitution
\(x-1=t^{2},\)
what represents the true evaluated reverse-substituted algebraic form?

A \(2(x-1)^{2}+3x+C\)

B \(\frac{2}{3}(x-1)^{3/2}+2\sqrt{x-1}+C\)

C \(log⁡∣x-1∣+C\)

D \(\frac{1}{3}(x-1)+(x-1)^{2}+C\)

💡 Short Explanation

Let \(x-1=t^{2}\). Rewrite the integral completely in terms of \(t\). Integrate and substitute back.

✅ Correct Answer Explanation

Using \(x-1=t^{2},\) we have \(x=t^{2}+1,dx=2t dt,\sqrt{x-1}=t.\) Therefore, \(\int \frac{x}{\sqrt{x-1}}dx=2\int (t^{2}+1) dt.\) Integrating, \(2\left(\frac{t^{3}}{3},\ t\right)+C.\) Replacing \(t=\sqrt{x-1},\) gives \(\frac{2}{3}(x-1)^{3/2}+2\sqrt{x-1}+C.\) Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Does not result from the substitution method.
Option C → The integral is not of the logarithmic form \(\frac{f^{'}(x)}{f(x)}\).
Option D → Incorrectly integrates the transformed polynomial.

🎯 Strategy

Used

Substitution

Application:

Replace the radical with a polynomial expression and integrate.

Final Logic:

The substitution converts the radical integral into a polynomial integral.

🧠 Memory Trick

"Root → Square Variable → Integrate → Substitute Back."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Integration by Radical Substitution

📄 Passage

Definite integrals play a highly crucial calculation role in economic optimization frameworks, such as deriving total revenue from marginal revenue models. To accurately find specific economic volumes, one must often formally integrate composite or exponential functions where structured substitution techniques drastically simplify the mechanical workload. Transforming limits directly during these definite integrals is a cornerstone technique because it entirely avoids the complex algebraic backtracking of indefinite reverse substitution, thereby leading straight to the final numerical evaluation.

According to the passage, what represents the primary mathematical advantage of directly transforming integration limits during numerical definite integrals?

A It fundamentally ignores the arbitrary mathematical constant \(C\) globally in all possible economic models.

B It permanently modifies the non-linear demand curve into a linear baseline function automatically.

C It strictly leads straight to a numerical value, avoiding the tedious algebraic tracking of re-substituting \(x\).

D It numerically guarantees the financial surplus output is always a universally positive prime integer.

💡 Short Explanation

The passage discusses substitution in definite integrals. Changing limits removes the need for reverse substitution. This makes evaluation quicker and simpler.

✅ Correct Answer Explanation

For definite integrals, the limits are transformed into the new variable after substitution. This allows the integral to be evaluated directly without converting back to the original variable. The passage explicitly states that this technique avoids the complex algebraic backtracking of reverse substitution, leading directly to the final numerical value. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → The constant of integration is naturally absent in definite integrals, but changing limits is not the reason.
Option B → Substitution does not alter the nature of demand curves.
Option D → No mathematical method guarantees a positive prime result.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Locate the statement in the passage describing the advantage of transforming limits.

Final Logic:

The passage directly states that changing limits avoids reverse substitution.

🧠 Memory Trick

"New Limits → No Back Substitution."

📚 Class XII – Mathematics Part II – Chapter: Definite Integrals – Topic: Change of Limits in Substitution

📄 Passage

If a company's Marginal Cost is precisely provided as
\(MC=\frac{x}{\sqrt{2500+x^{2}}},\)
and fixed initial costs are ₹1000, what determines the explicit final Cost function \(C(x)\)?

A \(C(x)=\sqrt{2500+x^{2}}+1000\)

B \(C(x)=\frac{1}{\sqrt{2500+x^{2}}}+1000\)

C \(C(x)=x\sqrt{2500+x^{2}}+950\)

D \(C(x)=\sqrt{2500+x^{2}}+950\)

💡 Short Explanation

Integrate the marginal cost. Use the initial condition \(C(0)=1000\). Determine the constant of integration.

✅ Correct Answer Explanation

Integrate \(\frac{x}{\sqrt{2500+x^{2}}}.\) Let \(t=2500+x^{2}.\) Then, \(dt=2x dx.\) Hence, \(C(x)=\sqrt{2500+x^{2}}+C.\) Using \(C(0)=1000,\) we get \(50+C=1000,\) so \(C=950.\) Therefore, \(C(x)=\sqrt{2500+x^{2}}+950.\) Hence, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Gives \(C(0)=1050\), not ₹1000.
Option B → Is not the antiderivative of the marginal cost.
Option C → Incorrect antiderivative.

🎯 Strategy

Used

Substitution

Application:

Integrate the marginal cost and use the initial condition to determine the constant.

Final Logic:

Integration followed by the fixed-cost condition gives Option D.

🧠 Memory Trick

"Integrate MC → Use Fixed Cost."

📚 Class XII – Mathematics Part II – Chapter: Applications of Integrals – Topic: Marginal Cost and Total Cost Functions

19 Evaluate the following continuous radical fractional substitution:
\(\int \frac{1-2x}{\sqrt{1+x^{2}}} dx\)
by deliberately splitting the term into two separate integrals.

A \(\frac{1}{\sqrt{1+x^{2}}}-2x+C\)

B \(log⁡∣x+\sqrt{1+x^{2}}∣+2\sqrt{1+x^{2}}+C\)

C \(log⁡∣x+\sqrt{1+x^{2}}∣-2\sqrt{1+x^{2}}+C\)

D \(log⁡∣1+x^{2}∣-\sqrt{1+x^{2}}+C\)

💡 Short Explanation

Split the integral into two simpler integrals. Apply the standard logarithmic integral to the first part. Use substitution for the second part.

✅ Correct Answer Explanation

Split the integral: \(\int \frac{1-2x}{\sqrt{1+x^{2}}} dx=\int \frac{dx}{\sqrt{1+x^{2}}}-2\int \frac{x}{\sqrt{1+x^{2}}} dx.\) For the first integral, \(\int \frac{dx}{\sqrt{1+x^{2}}}=log⁡∣x+\sqrt{1+x^{2}}∣.\) For the second, let \(t=1+x^{2},\) so \(dt=2x dx.\) Hence, \(2\int \frac{x}{\sqrt{1+x^{2}}}dx=2\sqrt{1+x^{2}}.\) Therefore, \(log⁡∣x+\sqrt{1+x^{2}}∣-2\sqrt{1+x^{2}}+C.\) Thus, Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Does not result from integrating either term correctly.
Option B → Uses the wrong sign before the radical term.
Option D → Applies an incorrect logarithmic identity.

🎯 Strategy

Used

Substitution

Application:

Split the integral and evaluate each part using the appropriate standard method.

Final Logic:

One part gives a logarithm, while the other gives a radical term with a negative sign.

🧠 Memory Trick

"Split → Log + Root."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Integration by Splitting and Substitution

20 Determine the finalized quadratic integral by applying completing the square techniques:
\(\int \frac{1}{\sqrt{3x^{2}+2x-1}} dx.\)

A \(log⁡∣x+\frac{1}{3}+\sqrt{3x^{2}+2x-1}∣+C\)

B \(\frac{1}{3}log⁡∣x+\sqrt{x^{2}+2x-1}∣+C\)

C \(\frac{1}{\sqrt{3}}{sin⁡}^{-1}\left(\frac{x}{3}\right)+C\)

D \(\frac{1}{\sqrt{3}}log⁡∣x+\frac{1}{3}+\sqrt{x^{2}+\frac{2}{3}x-\frac{1}{3}}∣+C\)

💡 Short Explanation

Complete the square for the quadratic expression. Convert the integral into a standard logarithmic form. Apply the standard integral formula.

✅ Correct Answer Explanation

First factor the quadratic: \(3x^{2}+2x-1=3\left(x^{2}+\frac{2}{3}x-\frac{1}{3}\right).\) Hence, \(\sqrt{3x^{2}+2x-1}=\sqrt{3}\sqrt{x^{2}+\frac{2}{3}x-\frac{1}{3}}.\) Therefore, \(\int \frac{dx}{\sqrt{3x^{2}+2x-1}}=\frac{1}{\sqrt{3}}\int \frac{dx}{\sqrt{x^{2}+\frac{2}{3}x-\frac{1}{3}}}.\) Applying the standard logarithmic result, \(\frac{1}{\sqrt{3}}log⁡∣x+\frac{1}{3}+\sqrt{x^{2}+\frac{2}{3}x-\frac{1}{3}}∣+C.\) Hence, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Omits the factor \(\frac{1}{\sqrt{3}}\).
Option B → Uses an incorrect coefficient and quadratic expression.
Option C → The inverse sine formula applies to \(\sqrt{a^{2}-x^{2}}\), not the given expression.

🎯 Strategy

Used

Substitution

Application:

Complete the square and convert the integral into the corresponding standard logarithmic form.

Final Logic:

After completing the square, the standard logarithmic integral gives Option D.

🧠 Memory Trick

"Complete the Square → Log Formula."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Standard Integrals Using Completing the Square

CUET UG Applied Mathematics Booster Test 2 - Integration by Substitution

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