JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Definite integration using algebraic substitution and partial fractions

The integral 16 integral from 1 to 2 of dx/(x^3(x^2+2)^2) is equal to

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A11/12 - log_e 4B11/6 - log_e 4C11/6 + log_e 4D11/12 + log_e 4

Step-by-step solutionView

Correct answer

Using the substitution $t = x^{2}$ followed by partial fraction decomposition, the integral evaluates to $\frac{11}{6} - \log_{e} 4$ .

Option analysis

Why each option works or fails

A · 11/12 - log_e 4

Forgetting to distribute the overall coefficient 16 or halving the rational component incorrectly during evaluation. Carefully track overall multipliers when integrating each decomposed term, ensuring the factor of $16$ multiplies the entire integrated expression.

B · 11/6 - log_e 4

This is the correct evaluation of the definite integral. Correctly applying $t = x^{2}$ , decomposing into partial fractions, and evaluating between $t = 1$ and $t = 4$ yields $\frac{11}{6} - \log_{e} 4$ .

C · 11/6 + log_e 4

Reversing the signs of the logarithmic terms during partial fraction integration, leading to $+ \log_{e} (t / (t + 2))$ evaluated backwards. Ensure that $\int (\frac{1}{t} - \frac{1}{t + 2}) d t = \ln ∣ t ∣ - \ln ∣ t + 2 ∣$ preserves the minus sign on the second term.

D · 11/12 + log_e 4

Combining an arithmetic halving error with an incorrect sign on the logarithmic term. Keep logarithmic signs consistent when applying the fundamental theorem of calculus and double-check common denominators.

Reviewed route

Solution

StepWorking

01Given

Integral $I = 16 \int_{1}^{2} \frac{d x}{x^{3} (x^{2} + 2)^{2}}$ .

02Goal

Evaluate the definite integral in closed form.

03Approach

Factor out $x^{4}$ from $(x^{2} + 2)^{2}$ to write the denominator as $x^{7} (1 + 2 / x^{2})^{2}$ , then substitute $t = 1 + \frac{2}{x^{2}}$ .

04Execute

Rewrite the integrand: $I = 16 \int_{1}^{2} \frac{d x}{x^{7} {(1 + \frac{2}{x^{2}})}^{2}}$ . Let $t = 1 + \frac{2}{x^{2}}$ , so $d t = - \frac{4}{x^{3}} d x$ , giving $\frac{d x}{x^{3}} = - \frac{d t}{4}$ . Also, $\frac{2}{x^{2}} = t - 1 ⟹ \frac{1}{x^{4}} = \frac{(t - 1)^{2}}{4}$ . Thus, $\frac{d x}{x^{7}} = \frac{1}{x^{4}} \cdot \frac{d x}{x^{3}} = \frac{(t - 1)^{2}}{4} (- \frac{d t}{4}) = - \frac{(t - 1)^{2}}{16} d t$ . Limits: at $x = 1$ , $t = 3$ ; at $x = 2$ , $t = 1 + 2 / 4 = 3 / 2$ .

05Execute

Substitute into the integral: $I = 16 \int_{3}^{3 / 2} \frac{1}{t^{2}} (- \frac{(t - 1)^{2}}{16}) d t = \int_{3 / 2}^{3} \frac{t^{2} - 2 t + 1}{t^{2}} d t = \int_{3 / 2}^{3} (1 - \frac{2}{t} + \frac{1}{t^{2}}) d t$ .

06Execute

Integrate term by term: ${[t - 2 \ln t - \frac{1}{t}]}_{3 / 2}^{3} = (3 - \frac{3}{2}) - 2 (\ln 3 - \ln (3 / 2)) - (\frac{1}{3} - \frac{2}{3}) = \frac{3}{2} - 2 \ln 2 + \frac{1}{3} = \frac{11}{6} - \ln 4$ .

✓Verify

Numerical sanity check: $11 / 6 - \ln 4 \approx 1.833 - 1.386 = 0.447 > 0$ . Over $[1, 2]$ , the integrand $16 / (x^{3} (x^{2} + 2)^{2})$ ranges from $16 / (1 \cdot 9) \approx 1.78$ down to $16 / (8 \cdot 36) \approx 0.056$ , with an average roughly around $0.45$ , matching the result.

Hints that build this answer step by step

What is the most effective substitution to simplify the integrand $\frac{1}{x^{3} (x^{2} + 2)^{2}}$ ?
Multiply numerator and denominator by $x$ , then substitute $t = x^{2}$ with $d t = 2 x d x$ .
With $I = 16 \int_{1}^{2} \frac{x d x}{x^{4} (x^{2} + 2)^{2}}$ and $t = x^{2}$ , what is the transformed definite integral in terms of $t$ ?
$8 \int_{1}^{4} \frac{d t}{t^{2} (t + 2)^{2}}$
What is the partial fraction decomposition of $\frac{8}{t^{2} (t + 2)^{2}}$ ?
$- \frac{1}{t} + \frac{2}{t^{2}} + \frac{1}{t + 2} + \frac{2}{(t + 2)^{2}}$
Evaluating the antiderivative ${[\ln ∣ \frac{t + 2}{t} ∣ - \frac{2}{t} - \frac{2}{t + 2}]}_{1}^{4}$ , what is the final value?
$\frac{11}{6} - \log_{e} 4$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why pull out x^2 from the bracket instead of using partial fractions directly?

Factoring out x^2 creates negative powers of x whose derivative directly matches 1/x^3, avoiding tedious high-degree partial fractions.

Answer

Using the substitution $t = x^{2}$ followed by partial fraction decomposition, the integral evaluates to $\frac{11}{6} - \log_{e} 4$ .

Why each option works or fails

A: 11/12 - log_e 4 - Forgetting to distribute the overall coefficient 16 or halving the rational component incorrectly during evaluation. Carefully track overall multipliers when integrating each decomposed term, ensuring the factor of $16$ multiplies the entire integrated expression.
B · correct: 11/6 - log_e 4 - This is the correct evaluation of the definite integral. Correctly applying $t = x^{2}$ , decomposing into partial fractions, and evaluating between $t = 1$ and $t = 4$ yields $\frac{11}{6} - \log_{e} 4$ .
C: 11/6 + log_e 4 - Reversing the signs of the logarithmic terms during partial fraction integration, leading to $+ \log_{e} (t / (t + 2))$ evaluated backwards. Ensure that $\int (\frac{1}{t} - \frac{1}{t + 2}) d t = \ln ∣ t ∣ - \ln ∣ t + 2 ∣$ preserves the minus sign on the second term.
D: 11/12 + log_e 4 - Combining an arithmetic halving error with an incorrect sign on the logarithmic term. Keep logarithmic signs consistent when applying the fundamental theorem of calculus and double-check common denominators.

Step-by-step solution

Given: Integral $I = 16 \int_{1}^{2} \frac{d x}{x^{3} (x^{2} + 2)^{2}}$ .
Goal: Evaluate the definite integral in closed form.
Approach: Factor out $x^{4}$ from $(x^{2} + 2)^{2}$ to write the denominator as $x^{7} (1 + 2 / x^{2})^{2}$ , then substitute $t = 1 + \frac{2}{x^{2}}$ .
Execute: Rewrite the integrand: $I = 16 \int_{1}^{2} \frac{d x}{x^{7} {(1 + \frac{2}{x^{2}})}^{2}}$ . Let $t = 1 + \frac{2}{x^{2}}$ , so $d t = - \frac{4}{x^{3}} d x$ , giving $\frac{d x}{x^{3}} = - \frac{d t}{4}$ . Also, $\frac{2}{x^{2}} = t - 1 ⟹ \frac{1}{x^{4}} = \frac{(t - 1)^{2}}{4}$ . Thus, $\frac{d x}{x^{7}} = \frac{1}{x^{4}} \cdot \frac{d x}{x^{3}} = \frac{(t - 1)^{2}}{4} (- \frac{d t}{4}) = - \frac{(t - 1)^{2}}{16} d t$ . Limits: at $x = 1$ , $t = 3$ ; at $x = 2$ , $t = 1 + 2 / 4 = 3 / 2$ .
Execute: Substitute into the integral: $I = 16 \int_{3}^{3 / 2} \frac{1}{t^{2}} (- \frac{(t - 1)^{2}}{16}) d t = \int_{3 / 2}^{3} \frac{t^{2} - 2 t + 1}{t^{2}} d t = \int_{3 / 2}^{3} (1 - \frac{2}{t} + \frac{1}{t^{2}}) d t$ .
Execute: Integrate term by term: ${[t - 2 \ln t - \frac{1}{t}]}_{3 / 2}^{3} = (3 - \frac{3}{2}) - 2 (\ln 3 - \ln (3 / 2)) - (\frac{1}{3} - \frac{2}{3}) = \frac{3}{2} - 2 \ln 2 + \frac{1}{3} = \frac{11}{6} - \ln 4$ .
Verify: Numerical sanity check: $11 / 6 - \ln 4 \approx 1.833 - 1.386 = 0.447 > 0$ . Over $[1, 2]$ , the integrand $16 / (x^{3} (x^{2} + 2)^{2})$ ranges from $16 / (1 \cdot 9) \approx 1.78$ down to $16 / (8 \cdot 36) \approx 0.056$ , with an average roughly around $0.45$ , matching the result.

Definite integration using algebraic substitution and partial fractions

What feels right?

Using the substitution $t = x^{2}$ followed by partial fraction decomposition, the integral evaluates to $\frac{11}{6} - \log_{e} 4$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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