JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Indefinite Integration Using Substitution and Partial Fractions

Q: Why is the absolute value removed in ln|t+1|?

Because t = x^2 ≥ 0, both t+1 and t+3 are strictly positive for all real x.

Let f(x) = integral of 2x/((x^2+1)(x^2+3)) dx. If f(3) = 1/2(log_e 5 - log_e 6), then f(4) is equal to

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Alog_e 19 - log_e 20Blog_e 17 - log_e 18C1/2(log_e 19 - log_e 17)D1/2(log_e 17 - log_e 19)

Step-by-step solutionView

Correct answer

Using the substitution $t = x^{2}$ and partial fractions gives $f (x) = \frac{1}{2} (\log_{e} (x^{2} + 1) - \log_{e} (x^{2} + 3)) + C$ ; with $C = 0$ determined from $f (3)$ , evaluating at $x = 4$ yields $\frac{1}{2} (\log_{e} 17 - \log_{e} 19)$ .

Option analysis

Why each option works or fails

A · log_e 19 - log_e 20

Omitting the factor of $\frac{1}{2}$ from the partial fraction decomposition $\frac{1}{(t + 1) (t + 3)} = \frac{1}{2} (\frac{1}{t + 1} - \frac{1}{t + 3})$ and substituting $x + 3$ or miscomputing $4^{2} + 3$ . Ensure the constants in partial fractions are computed correctly by checking common denominators, and evaluate $x^{2} + 1$ and $x^{2} + 3$ accurately at $x = 4$ .

B · log_e 17 - log_e 18

Forgetting the scalar coefficient $\frac{1}{2}$ in the partial fraction expansion of $\frac{1}{(t + 1) (t + 3)}$ . When decomposing $\frac{1}{(t + 1) (t + 3)}$ , note that $(t + 3) - (t + 1) = 2$ , so multiply the difference of terms by $\frac{1}{2}$ .

C · 1/2(log_e 19 - log_e 17)

Reversing the signs in the partial fraction decomposition, writing $\frac{1}{t + 3} - \frac{1}{t + 1}$ instead of $\frac{1}{t + 1} - \frac{1}{t + 3}$ . Check the sign by recombining terms: $\frac{1}{t + 1} - \frac{1}{t + 3} = \frac{(t + 3) - (t + 1)}{(t + 1) (t + 3)} = \frac{2}{(t + 1) (t + 3)}$ .

D · 1/2(log_e 17 - log_e 19)

This is the correct evaluation of $f (4)$ . Correctly integrated using $t = x^{2}$ , found $C = 0$ via $f (3)$ , and computed $f (4) = \frac{1}{2} (\log_{e} 17 - \log_{e} 19)$ .

Reviewed route

Solution

StepWorking

01Given

$f (x) = \int \frac{2 x}{(x^{2} + 1) (x^{2} + 3)} d x$ and $f (3) = \frac{1}{2} (\log_{e} 5 - \log_{e} 6)$ .

02Goal

Find the value of $f (4)$ .

03Approach

Substitute $t = x^{2}$ so that $d t = 2 x d x$ . Then resolve into partial fractions: $\frac{1}{(t + 1) (t + 3)} = \frac{1}{2} (\frac{1}{t + 1} - \frac{1}{t + 3})$ . Integrate to find $f (x)$ with constant $C$ , determine $C$ using $f (3)$ , and evaluate $f (4)$ .

04Execute

Substitute $t = x^{2}$ , $d t = 2 x d x$ : $f (x) = \int \frac{d t}{(t + 1) (t + 3)} = \frac{1}{2} \int (\frac{1}{t + 1} - \frac{1}{t + 3}) d t = \frac{1}{2} \ln ∣ \frac{t + 1}{t + 3} ∣ + C = \frac{1}{2} \ln (\frac{x^{2} + 1}{x^{2} + 3}) + C$ Using $f (3)$ : $f (3) = \frac{1}{2} \ln (\frac{3^{2} + 1}{3^{2} + 3}) + C = \frac{1}{2} \ln (\frac{10}{12}) + C = \frac{1}{2} \ln (\frac{5}{6}) + C$ Given $f (3) = \frac{1}{2} (\log_{e} 5 - \log_{e} 6) = \frac{1}{2} \ln (\frac{5}{6})$ , we find $C = 0$ . Now evaluate at $x = 4$ : $f (4) = \frac{1}{2} \ln (\frac{4^{2} + 1}{4^{2} + 3}) = \frac{1}{2} \ln (\frac{17}{19}) = \frac{1}{2} (\log_{e} 17 - \log_{e} 19)$

✓Verify

Check limits and signs: for $x > 0$ , $\frac{x^{2} + 1}{x^{2} + 3} < 1$ , so $\ln (\frac{x^{2} + 1}{x^{2} + 3}) < 0$ . Thus $f (4) = \frac{1}{2} (\log_{e} 17 - \log_{e} 19) < 0$ , which is consistent since $17 < 19$ .

Hints that build this answer step by step

What substitution simplifies the integrand $\frac{2 x}{(x^{2} + 1) (x^{2} + 3)}$ ?
Substitute $t = x^{2}$ , so $d t = 2 x d x$
How does $\int \frac{d t}{(t + 1) (t + 3)}$ decompose into partial fractions?
$\frac{1}{2} \int (\frac{1}{t + 1} - \frac{1}{t + 3}) d t$
Given $f (x) = \frac{1}{2} \log_{e} (\frac{x^{2} + 1}{x^{2} + 3}) + C$ and $f (3) = \frac{1}{2} (\log_{e} 5 - \log_{e} 6)$ , what is the value of $C$ ?
$C = 0$
With $f (x) = \frac{1}{2} (\log_{e} (x^{2} + 1) - \log_{e} (x^{2} + 3))$ , what is $f (4)$ ?
$\frac{1}{2} (\log_{e} 17 - \log_{e} 19)$

✓ 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: 8 September 2026

Quick checks

Students also ask

Why is the absolute value removed in

\ln ∣ t + 1 ∣

Because $t = x^{2} \geq 0$ , both $t + 1$ and $t + 3$ are strictly positive for all real $x$ .

Answer

Using the substitution $t = x^{2}$ and partial fractions gives $f (x) = \frac{1}{2} (\log_{e} (x^{2} + 1) - \log_{e} (x^{2} + 3)) + C$ ; with $C = 0$ determined from $f (3)$ , evaluating at $x = 4$ yields $\frac{1}{2} (\log_{e} 17 - \log_{e} 19)$ .

Why each option works or fails

A: log_e 19 - log_e 20 - Omitting the factor of $\frac{1}{2}$ from the partial fraction decomposition $\frac{1}{(t + 1) (t + 3)} = \frac{1}{2} (\frac{1}{t + 1} - \frac{1}{t + 3})$ and substituting $x + 3$ or miscomputing $4^{2} + 3$ . Ensure the constants in partial fractions are computed correctly by checking common denominators, and evaluate $x^{2} + 1$ and $x^{2} + 3$ accurately at $x = 4$ .
B: log_e 17 - log_e 18 - Forgetting the scalar coefficient $\frac{1}{2}$ in the partial fraction expansion of $\frac{1}{(t + 1) (t + 3)}$ . When decomposing $\frac{1}{(t + 1) (t + 3)}$ , note that $(t + 3) - (t + 1) = 2$ , so multiply the difference of terms by $\frac{1}{2}$ .
C: 1/2(log_e 19 - log_e 17) - Reversing the signs in the partial fraction decomposition, writing $\frac{1}{t + 3} - \frac{1}{t + 1}$ instead of $\frac{1}{t + 1} - \frac{1}{t + 3}$ . Check the sign by recombining terms: $\frac{1}{t + 1} - \frac{1}{t + 3} = \frac{(t + 3) - (t + 1)}{(t + 1) (t + 3)} = \frac{2}{(t + 1) (t + 3)}$ .
D · correct: 1/2(log_e 17 - log_e 19) - This is the correct evaluation of $f (4)$ . Correctly integrated using $t = x^{2}$ , found $C = 0$ via $f (3)$ , and computed $f (4) = \frac{1}{2} (\log_{e} 17 - \log_{e} 19)$ .

Step-by-step solution

Given: $f (x) = \int \frac{2 x}{(x^{2} + 1) (x^{2} + 3)} d x$ and $f (3) = \frac{1}{2} (\log_{e} 5 - \log_{e} 6)$ .
Goal: Find the value of $f (4)$ .
Approach: Substitute $t = x^{2}$ so that $d t = 2 x d x$ . Then resolve into partial fractions: $\frac{1}{(t + 1) (t + 3)} = \frac{1}{2} (\frac{1}{t + 1} - \frac{1}{t + 3})$ . Integrate to find $f (x)$ with constant $C$ , determine $C$ using $f (3)$ , and evaluate $f (4)$ .
Execute: Substitute $t = x^{2}$ , $d t = 2 x d x$ : $f (x) = \int \frac{d t}{(t + 1) (t + 3)} = \frac{1}{2} \int (\frac{1}{t + 1} - \frac{1}{t + 3}) d t = \frac{1}{2} \ln ∣ \frac{t + 1}{t + 3} ∣ + C = \frac{1}{2} \ln (\frac{x^{2} + 1}{x^{2} + 3}) + C$ Using $f (3)$ : $f (3) = \frac{1}{2} \ln (\frac{3^{2} + 1}{3^{2} + 3}) + C = \frac{1}{2} \ln (\frac{10}{12}) + C = \frac{1}{2} \ln (\frac{5}{6}) + C$ Given $f (3) = \frac{1}{2} (\log_{e} 5 - \log_{e} 6) = \frac{1}{2} \ln (\frac{5}{6})$ , we find $C = 0$ . Now evaluate at $x = 4$ : $f (4) = \frac{1}{2} \ln (\frac{4^{2} + 1}{4^{2} + 3}) = \frac{1}{2} \ln (\frac{17}{19}) = \frac{1}{2} (\log_{e} 17 - \log_{e} 19)$
Verify: Check limits and signs: for $x > 0$ , $\frac{x^{2} + 1}{x^{2} + 3} < 1$ , so $\ln (\frac{x^{2} + 1}{x^{2} + 3}) < 0$ . Thus $f (4) = \frac{1}{2} (\log_{e} 17 - \log_{e} 19) < 0$ , which is consistent since $17 < 19$ .

Indefinite Integration Using Substitution and Partial Fractions

What feels right?

Using the substitution $t = x^{2}$ and partial fractions gives $f (x) = \frac{1}{2} (\log_{e} (x^{2} + 1) - \log_{e} (x^{2} + 3)) + C$ ; with $C = 0$ determined from $f (3)$ , evaluating at $x = 4$ yields $\frac{1}{2} (\log_{e} 17 - \log_{e} 19)$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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