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Integral Calculus: JEE Main Mathematics Question with Solution

The integral

16 \int_1^2 \frac{dx}{x^3(x^2+2)^2}

is equal to

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\frac{11}{12} - \log_e 4

\frac{11}{6} - \log_e 4

\frac{11}{6} + \log_e 4

\frac{11}{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 ·

\frac{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 ·

\frac{11}{6} - \log_e 4

None. 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 ·

\frac{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 $\int (\frac{1}{t} - \frac{1}{t+2}) dt = \ln|t| - \ln|t+2|$ preserves the minus sign on the second term.

D ·

\frac{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 $I = 16 \int_1^2 \frac{dx}{x^3(x^2+2)^2}$ .

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: $I = 16 \int_1^2 \frac{dx}{x^7 \left(1 + \frac{2}{x^2}\right)^2}$ . Let $t = 1 + \frac{2}{x^2}$ , so $dt = -\frac{4}{x^3} dx$ , giving $\frac{dx}{x^3} = -\frac{dt}{4}$ . Also, $\frac{2}{x^2} = t - 1 \implies \frac{1}{x^4} = \frac{(t-1)^2}{4}$ . Thus, $\frac{dx}{x^7} = \frac{1}{x^4} \cdot \frac{dx}{x^3} = \frac{(t-1)^2}{4} \left(-\frac{dt}{4}\right) = -\frac{(t-1)^2}{16} dt$ . Limits: at $x = 1$ , $t = 3$ ; at $x = 2$ , $t = 1 + 2/4 = 3/2$ .

05execute

Substitute into the integral: $I = 16 \int_3^{3/2} \frac{1}{t^2} \left(-\frac{(t-1)^2}{16}\right) dt = \int_{3/2}^3 \frac{t^2 - 2t + 1}{t^2} dt = \int_{3/2}^3 \left(1 - \frac{2}{t} + \frac{1}{t^2}\right) dt$ .

06execute

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

✓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 $dt = 2x\,dx$ .
With $I = 16 \int_1^2 \frac{x\,dx}{x^4(x^2+2)^2}$ and $t = x^2$ , what is the transformed definite integral in terms of $t$ ?
$8 \int_1^4 \frac{dt}{t^2(t+2)^2}$
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 $\left[ \ln\left|\frac{t+2}{t}\right| - \frac{2}{t} - \frac{2}{t+2} \right]_1^4$ , what is the final value?
$\frac{11}{6} - \log_e 4$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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: $\frac{11}{12} - \log_e 4$ - 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: $\frac{11}{6} - \log_e 4$ - None. 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: $\frac{11}{6} + \log_e 4$ - Reversing the signs of the logarithmic terms during partial fraction integration, leading to $+\log_e(t/(t+2))$ evaluated backwards. Ensure that $\int (\frac{1}{t} - \frac{1}{t+2}) dt = \ln|t| - \ln|t+2|$ preserves the minus sign on the second term.
D: $\frac{11}{12} + \log_e 4$ - 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 $I = 16 \int_1^2 \frac{dx}{x^3(x^2+2)^2}$ .
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: $I = 16 \int_1^2 \frac{dx}{x^7 \left(1 + \frac{2}{x^2}\right)^2}$ . Let $t = 1 + \frac{2}{x^2}$ , so $dt = -\frac{4}{x^3} dx$ , giving $\frac{dx}{x^3} = -\frac{dt}{4}$ . Also, $\frac{2}{x^2} = t - 1 \implies \frac{1}{x^4} = \frac{(t-1)^2}{4}$ . Thus, $\frac{dx}{x^7} = \frac{1}{x^4} \cdot \frac{dx}{x^3} = \frac{(t-1)^2}{4} \left(-\frac{dt}{4}\right) = -\frac{(t-1)^2}{16} dt$ . Limits: at $x = 1$ , $t = 3$ ; at $x = 2$ , $t = 1 + 2/4 = 3/2$ .
execute: Substitute into the integral: $I = 16 \int_3^{3/2} \frac{1}{t^2} \left(-\frac{(t-1)^2}{16}\right) dt = \int_{3/2}^3 \frac{t^2 - 2t + 1}{t^2} dt = \int_{3/2}^3 \left(1 - \frac{2}{t} + \frac{1}{t^2}\right) dt$ .
execute: Integrate term by term: $\frac{dx}{x^7} = \frac{1}{x^4} \cdot \frac{dx}{x^3} = \frac{(t-1)^2}{4} \left(-\frac{dt}{4}\right) = -\frac{(t-1)^2}{16} dt$ .
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.

Integral Calculus: JEE Main Mathematics Question with Solution

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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