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Integral Calculus: JEE Main Mathematics Question with Solution

Q: Why is the absolute value removed in ?

Because , both and are strictly positive for all real .

Let

f(x) = \int \frac{2x}{(x^2+1)(x^2+3)} dx

. If

f(3) = \frac{1}{2}(\log_e 5 - \log_e 6)

, then

f(4)

is equal to

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What feels right?

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\log_e 19 - \log_e 20

\log_e 17 - \log_e 18

\frac{1}{2}(\log_e 19 - \log_e 17)

\frac{1}{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 $\frac{1}{(t+1)(t+3)} = \frac{1}{2}\left(\frac{1}{t+1} - \frac{1}{t+3}\right)$ 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 ·

\frac{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 ·

\frac{1}{2}(\log_e 17 - \log_e 19)

None. 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

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

04execute

Substitute $t = x^2$ , $dt = 2x\,dx$ : $f(x) = \int \frac{dt}{(t+1)(t+3)} = \frac{1}{2} \int \left(\frac{1}{t+1} - \frac{1}{t+3}\right) dt = \frac{1}{2} \ln\left|\frac{t+1}{t+3}\right| + C = \frac{1}{2} \ln\left(\frac{x^2+1}{x^2+3}\right) + C$ Using $f(3)$ : $f(3) = \frac{1}{2} \ln\left(\frac{3^2+1}{3^2+3}\right) + C = \frac{1}{2} \ln\left(\frac{10}{12}\right) + C = \frac{1}{2} \ln\left(\frac{5}{6}\right) + C$ Given $f(3) = \frac{1}{2}(\log_e 5 - \log_e 6) = \frac{1}{2} \ln\left(\frac{5}{6}\right)$ , we find $C = 0$ . Now evaluate at $x = 4$ : $f(4) = \frac{1}{2} \ln\left(\frac{4^2+1}{4^2+3}\right) = \frac{1}{2} \ln\left(\frac{17}{19}\right) = \frac{1}{2}(\log_e 17 - \log_e 19)$

✓verify

Check limits and signs: for $x > 0$ , $\frac{x^2+1}{x^2+3} < 1$ , so $\ln\left(\frac{x^2+1}{x^2+3}\right) < 0$ . 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 $\frac{2x}{(x^2+1)(x^2+3)}$ ?
Substitute $t = x^2$ , so $dt = 2x\,dx$
How does $\int \frac{dt}{(t+1)(t+3)}$ decompose into partial fractions?
$\frac{1}{2} \int \left(\frac{1}{t+1} - \frac{1}{t+3}\right) dt$
Given $f(x) = \frac{1}{2}\log_e\left(\frac{x^2+1}{x^2+3}\right) + C$ 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
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why is the absolute value removed in

\ln|t+1|

Because $t = x^2 \ge 0$ , 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 $\frac{1}{(t+1)(t+3)} = \frac{1}{2}\left(\frac{1}{t+1} - \frac{1}{t+3}\right)$ 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: $\frac{1}{2}(\log_e 19 - \log_e 17)$ - 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: $\frac{1}{2}(\log_e 17 - \log_e 19)$ - None. 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: $f(x) = \int \frac{2x}{(x^2+1)(x^2+3)}\,dx$ 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 $dt = 2x\,dx$ . Then resolve into partial fractions: $\frac{1}{(t+1)(t+3)} = \frac{1}{2}\left(\frac{1}{t+1} - \frac{1}{t+3}\right)$ . Integrate to find $f(x)$ with constant $C$ , determine $C$ using $f(3)$ , and evaluate $f(4)$ .
execute: Substitute $t = x^2$ , $dt = 2x\,dx$ : $f(x) = \int \frac{dt}{(t+1)(t+3)} = \frac{1}{2} \int \left(\frac{1}{t+1} - \frac{1}{t+3}\right) dt = \frac{1}{2} \ln\left|\frac{t+1}{t+3}\right| + C = \frac{1}{2} \ln\left(\frac{x^2+1}{x^2+3}\right) + C$ Using $f(3)$ : $f(3) = \frac{1}{2} \ln\left(\frac{3^2+1}{3^2+3}\right) + C = \frac{1}{2} \ln\left(\frac{10}{12}\right) + C = \frac{1}{2} \ln\left(\frac{5}{6}\right) + C$ Given $f(3) = \frac{1}{2}(\log_e 5 - \log_e 6) = \frac{1}{2} \ln\left(\frac{5}{6}\right)$ , we find $C = 0$ . Now evaluate at $x = 4$ : $f(4) = \frac{1}{2} \ln\left(\frac{4^2+1}{4^2+3}\right) = \frac{1}{2} \ln\left(\frac{17}{19}\right) = \frac{1}{2}(\log_e 17 - \log_e 19)$
verify: Check limits and signs: for $x > 0$ , $\frac{x^2+1}{x^2+3} < 1$ , so $\ln\left(\frac{x^2+1}{x^2+3}\right) < 0$ . Thus $f(4) = \frac{1}{2}(\log_e 17 - \log_e 19) < 0$ , which is consistent since $17 < 19$ .

Integral Calculus: JEE Main Mathematics Question with Solution

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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