Why multiply by in both numerator and denominator?

Because the derivative of the inside term is . We already have in the numerator, so introducing completes the differential .

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Integral Calculus: JEE Main Mathematics Question with Solution

Let

\mathrm{I}(\mathrm{x})=\int \frac{(\mathrm{x}+1)}{\mathrm{x}\left(1+\mathrm{xe}^{\mathrm{x}}\right)^{2}} \mathrm{dx}, \mathrm{x}>0

, If

\lim _{\mathrm{x} \rightarrow \infty} \mathrm{I}(\mathrm{x})=0

, then

\mathrm{I}(1)

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.

A

\frac{\mathrm{e}+1}{\mathrm{e}+2}-\log _{\mathrm{e}}(\mathrm{e}+1)

B

\frac{\mathrm{e}+1}{\mathrm{e}+2}+\log _{\mathrm{e}}(\mathrm{e}+1)

C

\frac{\mathrm{e}+2}{\mathrm{e}+1}+\log _{\mathrm{e}}(\mathrm{e}+1)

D

\frac{\mathrm{e}+2}{\mathrm{e}+1}-\log _{\mathrm{e}}(\mathrm{e}+1)

Step-by-step solutionView

Correct answer

Multiplying the numerator and denominator by $e^x$ allows the substitution $t = x e^x$ , yielding $I(x) = \ln\left|\frac{x e^x}{1 + x e^x}\right| + \frac{1}{1 + x e^x} + 1$ , which gives $I(1) = \frac{e+2}{e+1} - \log_e(e+1)$ .

Option analysis

Why each option works or fails

A ·

\frac{\mathrm{e}+1}{\mathrm{e}+2}-\log _{\mathrm{e}}(\mathrm{e}+1)

Inverting the fraction $\frac{1}{1+e} + 1$ to get $\frac{e+1}{e+2}$ instead of $\frac{e+2}{e+1}$ . Carefully compute the sum $1 + \frac{1}{e+1} = \frac{e+1+1}{e+1} = \frac{e+2}{e+1}$ without inverting the numerator and denominator.

B ·

\frac{\mathrm{e}+1}{\mathrm{e}+2}+\log _{\mathrm{e}}(\mathrm{e}+1)

Changing the sign of the logarithm term during expansion and incorrectly combining the constant terms. Ensure $\ln\left(\frac{e}{1+e}\right) = \ln(e) - \ln(1+e) = 1 - \ln(1+e)$ , keeping the negative sign in front of the logarithm.

C ·

\frac{\mathrm{e}+2}{\mathrm{e}+1}+\log _{\mathrm{e}}(\mathrm{e}+1)

Flipping the sign of $-\log_e(e+1)$ into $+\log_e(e+1)$ when splitting $\ln\left(\frac{e}{1+e}\right)$ . Remember that $\ln(a/b) = \ln(a) - \ln(b)$ ; with $a=e$ and $b=e+1$ , this yields $1 - \ln(e+1)$ , not $1 + \ln(e+1)$ .

D ·

\frac{\mathrm{e}+2}{\mathrm{e}+1}-\log _{\mathrm{e}}(\mathrm{e}+1)

None. This option correctly evaluates the integral using $t = x e^x$ and uses the boundary condition $\lim_{x \to \infty} I(x) = 0$ to find $C = 0$ . None.

Reviewed route

Solution

StepWorking

01given

$I(x) = \int \frac{x+1}{x(1+xe^x)^2}\,dx$ for $x > 0$ , with $\lim_{x \to \infty} I(x) = 0$ .

02approach

Notice that $\frac{d}{dx}(x e^x) = (x+1)e^x$ . Multiply numerator and denominator by $e^x$ to create $(x+1)e^x\,dx$ in the numerator, then substitute $t = x e^x$ .

03execute

Multiply numerator and denominator by $e^x$ : $I(x) = \int \frac{(x+1)e^x}{xe^x(1+xe^x)^2}\,dx$ Let $t = xe^x \implies dt = (x+1)e^x\,dx$ . The integral transforms to: $\int \frac{dt}{t(1+t)^2}$

04execute

Decompose into partial fractions: $\frac{1}{t(1+t)^2} = \frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2}$ Integrating term-by-term: $I(x) = \ln|t| - \ln|1+t| + \frac{1}{1+t} + C = \ln\left(\frac{xe^x}{1+xe^x}\right) + \frac{1}{1+xe^x} + C$ Rewrite $\ln\left(\frac{xe^x}{1+xe^x}\right) = \ln\left(\frac{x e^x}{e^x(e^{-x}+x)}\right) = \ln\left(\frac{x}{x+e^{-x}}\right)$ .

05execute

Apply the boundary condition $\lim_{x\to\infty} I(x) = 0$ : $\lim_{x\to\infty} \left[\ln\left(\frac{x}{x+e^{-x}}\right) + \frac{1}{1+xe^x} + C\right] = \ln(1) + 0 + C = C \implies C = 0$ Therefore, $I(x) = \ln(xe^x) - \ln(1+xe^x) + \frac{1}{1+xe^x} = \ln x + x - \ln(1+xe^x) + \frac{1}{1+xe^x}$ .

06execute

Evaluate at $x = 1$ : $I(1) = \ln(1) + 1 - \ln(1+e) + \frac{1}{1+e} = 1 + \frac{1}{1+e} - \ln(1+e) = \frac{e+2}{e+1} - \ln(e+1)$

✓verify

Check that $1 + \frac{1}{e+1} = \frac{e+2}{e+1}$ , matching Option (3).

Hints that build this answer step by step

What algebraic manipulation makes the integrand $\frac{x+1}{x(1+xe^x)^2}$ directly integrable by substitution?
Multiply numerator and denominator by $e^x$ to produce $(x+1)e^x$ in the numerator and substitute $t = x e^x$ .
What is the partial fraction decomposition of $\frac{1}{t(1+t)^2}$ ?
$\frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2}$
Integrating $\int \left( \frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) dt$ gives $I(x) = \ln\left|\frac{t}{1+t}\right| + \frac{1}{1+t} + C$ . Given $\lim_{x \to \infty} I(x) = 0$ , what is $C$ ?
$C = 0$
With $I(x) = \ln\left(\frac{xe^x}{1+xe^x}\right) + \frac{1}{1+xe^x}$ , what is the value of $I(1)$ ?
$\frac{e+2}{e+1} - \log_e(e+1)$

✓ 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 multiply by

e^x

in both numerator and denominator?

Because the derivative of the inside term $x e^x$ is $(x+1)e^x$ . We already have $(x+1)$ in the numerator, so introducing $e^x$ completes the differential $d(xe^x)$ .

Answer

Multiplying the numerator and denominator by $e^x$ allows the substitution $t = x e^x$ , yielding $I(x) = \ln\left|\frac{x e^x}{1 + x e^x}\right| + \frac{1}{1 + x e^x} + 1$ , which gives $I(1) = \frac{e+2}{e+1} - \log_e(e+1)$ .

Why each option works or fails

A: $\frac{\mathrm{e}+1}{\mathrm{e}+2}-\log _{\mathrm{e}}(\mathrm{e}+1)$ - Inverting the fraction $\frac{1}{1+e} + 1$ to get $\frac{e+1}{e+2}$ instead of $\frac{e+2}{e+1}$ . Carefully compute the sum $1 + \frac{1}{e+1} = \frac{e+1+1}{e+1} = \frac{e+2}{e+1}$ without inverting the numerator and denominator.
B: $\frac{\mathrm{e}+1}{\mathrm{e}+2}+\log _{\mathrm{e}}(\mathrm{e}+1)$ - Changing the sign of the logarithm term during expansion and incorrectly combining the constant terms. Ensure $\ln\left(\frac{e}{1+e}\right) = \ln(e) - \ln(1+e) = 1 - \ln(1+e)$ , keeping the negative sign in front of the logarithm.
C: $\frac{\mathrm{e}+2}{\mathrm{e}+1}+\log _{\mathrm{e}}(\mathrm{e}+1)$ - Flipping the sign of $-\log_e(e+1)$ into $+\log_e(e+1)$ when splitting $\ln\left(\frac{e}{1+e}\right)$ . Remember that $\ln(a/b) = \ln(a) - \ln(b)$ ; with $a=e$ and $b=e+1$ , this yields $1 - \ln(e+1)$ , not $1 + \ln(e+1)$ .
D · correct: $\frac{\mathrm{e}+2}{\mathrm{e}+1}-\log _{\mathrm{e}}(\mathrm{e}+1)$ - None. This option correctly evaluates the integral using $t = x e^x$ and uses the boundary condition $\lim_{x \to \infty} I(x) = 0$ to find $C = 0$ . None.

Step-by-step solution

given: $I(x) = \int \frac{x+1}{x(1+xe^x)^2}\,dx$ for $x > 0$ , with $\lim_{x \to \infty} I(x) = 0$ .
approach: Notice that $\frac{d}{dx}(x e^x) = (x+1)e^x$ . Multiply numerator and denominator by $e^x$ to create $(x+1)e^x\,dx$ in the numerator, then substitute $t = x e^x$ .
execute: Multiply numerator and denominator by $e^x$ : $I(x) = \int \frac{(x+1)e^x}{xe^x(1+xe^x)^2}\,dx$ Let $t = xe^x \implies dt = (x+1)e^x\,dx$ . The integral transforms to: $\int \frac{dt}{t(1+t)^2}$
execute: Decompose into partial fractions: $\frac{1}{t(1+t)^2} = \frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2}$ Integrating term-by-term: $I(x) = \ln|t| - \ln|1+t| + \frac{1}{1+t} + C = \ln\left(\frac{xe^x}{1+xe^x}\right) + \frac{1}{1+xe^x} + C$ Rewrite $\ln\left(\frac{xe^x}{1+xe^x}\right) = \ln\left(\frac{x e^x}{e^x(e^{-x}+x)}\right) = \ln\left(\frac{x}{x+e^{-x}}\right)$ .
execute: Apply the boundary condition $\lim_{x\to\infty} I(x) = 0$ : $\lim_{x\to\infty} \left[\ln\left(\frac{x}{x+e^{-x}}\right) + \frac{1}{1+xe^x} + C\right] = \ln(1) + 0 + C = C \implies C = 0$ Therefore, $I(x) = \ln(xe^x) - \ln(1+xe^x) + \frac{1}{1+xe^x} = \ln x + x - \ln(1+xe^x) + \frac{1}{1+xe^x}$ .
execute: Evaluate at $x = 1$ : $I(1) = \ln(1) + 1 - \ln(1+e) + \frac{1}{1+e} = 1 + \frac{1}{1+e} - \ln(1+e) = \frac{e+2}{e+1} - \ln(e+1)$
verify: Check that $1 + \frac{1}{e+1} = \frac{e+2}{e+1}$ , matching Option (3).

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

Multiplying the numerator and denominator by $e^x$ allows the substitution $t = x e^x$ , yielding $I(x) = \ln\left|\frac{x e^x}{1 + x e^x}\right| + \frac{1}{1 + x e^x} + 1$ , which gives $I(1) = \frac{e+2}{e+1} - \log_e(e+1)$ .

Why each option works or fails

Solution

01given

02approach

03execute

04execute

05execute

06execute

✓verify

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