JEE MainMathematics

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Indefinite Integration via Substitution with Exponential Functions

Q: 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).

Let I(x)=integral of ((x+1))/(x(1+xe^x)^2) dx, x>0, If lim of _(x → ∞) I(x)=0, then I(1) is equal to

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A(e+1)/(e+2)-log _e(e+1)B(e+1)/(e+2)+log _e(e+1)C(e+2)/(e+1)+log _e(e+1)D(e+2)/(e+1)-log _e(e+1)

Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

A · (e+1)/(e+2)-log _e(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 · (e+1)/(e+2)+log _e(e+1)

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

C · (e+2)/(e+1)+log _e(e+1)

Flipping the sign of $- \log_{e} (e + 1)$ into $+ \log_{e} (e + 1)$ when splitting $\ln (\frac{e}{1 + e})$ . 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 · (e+2)/(e+1)-log _e(e+1)

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 $C = 0$ .

Reviewed route

Solution

StepWorking

01Given

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

02Approach

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

03Execute

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

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: $I (x) = \ln ∣ t ∣ - \ln ∣ 1 + t ∣ + \frac{1}{1 + t} + C = \ln (\frac{x e^{x}}{1 + x e^{x}}) + \frac{1}{1 + x e^{x}} + C$ Rewrite $\ln (\frac{x e^{x}}{1 + x e^{x}}) = \ln (\frac{x e^{x}}{e^{x} (e^{- x} + x)}) = \ln (\frac{x}{x + e^{- x}})$ .

05Execute

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

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 D.

Hints that build this answer step by step

What algebraic manipulation makes the integrand $\frac{x + 1}{x (1 + x e^{x})^{2}}$ 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 $\int (\frac{1}{t} - \frac{1}{1 + t} - \frac{1}{(1 + t)^{2}}) d t$ gives $I (x) = \ln ∣ \frac{t}{1 + t} ∣ + \frac{1}{1 + t} + C$ . Given $\lim_{x \to \infty} I (x) = 0$ , what is $C$ ?
$C = 0$
With $I (x) = \ln (\frac{x e^{x}}{1 + x e^{x}}) + \frac{1}{1 + x e^{x}}$ , 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
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $d (x e^{x})$ .

Answer

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

Why each option works or fails

A: (e+1)/(e+2)-log _e(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: (e+1)/(e+2)+log _e(e+1) - Changing the sign of the logarithm term during expansion and incorrectly combining the constant terms. Ensure $\ln (\frac{e}{1 + e}) = \ln (e) - \ln (1 + e) = 1 - \ln (1 + e)$ , keeping the negative sign in front of the logarithm.
C: (e+2)/(e+1)+log _e(e+1) - Flipping the sign of $- \log_{e} (e + 1)$ into $+ \log_{e} (e + 1)$ when splitting $\ln (\frac{e}{1 + e})$ . 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: (e+2)/(e+1)-log _e(e+1) - 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 $C = 0$ .

Step-by-step solution

Given: $I (x) = \int \frac{x + 1}{x (1 + x e^{x})^{2}} d x$ for $x > 0$ , with $\lim_{x \to \infty} I (x) = 0$ .
Approach: Notice that $\frac{d}{d x} (x e^{x}) = (x + 1) e^{x}$ . Multiply numerator and denominator by $e^{x}$ to create $(x + 1) e^{x} d x$ in the numerator, then substitute $t = x e^{x}$ .
Execute: Multiply numerator and denominator by $e^{x}$ : $I (x) = \int \frac{(x + 1) e^{x}}{x e^{x} (1 + x e^{x})^{2}} d x$ Let $t = x e^{x} ⟹ d t = (x + 1) e^{x} d x$ . The integral transforms to: $\int \frac{d t}{t (1 + t)^{2}}$
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: $I (x) = \ln ∣ t ∣ - \ln ∣ 1 + t ∣ + \frac{1}{1 + t} + C = \ln (\frac{x e^{x}}{1 + x e^{x}}) + \frac{1}{1 + x e^{x}} + C$ Rewrite $\ln (\frac{x e^{x}}{1 + x e^{x}}) = \ln (\frac{x e^{x}}{e^{x} (e^{- x} + x)}) = \ln (\frac{x}{x + e^{- x}})$ .
Execute: Apply the boundary condition $\lim_{x \to \infty} I (x) = 0$ : $\lim_{x \to \infty} [\ln (\frac{x}{x + e^{- x}}) + \frac{1}{1 + x e^{x}} + C] = \ln (1) + 0 + C = C ⟹ C = 0$ Therefore, $I (x) = \ln (x e^{x}) - \ln (1 + x e^{x}) + \frac{1}{1 + x e^{x}} = \ln x + x - \ln (1 + x e^{x}) + \frac{1}{1 + x e^{x}}$ .
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 D.

Indefinite Integration via Substitution with Exponential Functions

What feels right?

Multiplying the numerator and denominator by $e^{x}$ allows the substitution $t = x e^{x}$ , yielding $I (x) = \ln ∣ \frac{x e^{x}}{1 + x e^{x}} ∣ + \frac{1}{1 + x e^{x}} + 1$ , 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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Multiplying the numerator and denominator by exex allows the substitution t=xext=xex, yielding I(x)=ln⁡∣xex1+xex∣+11+xex+1I(x)=ln​1+xexxex​​+1+xex1​+1, which gives I(1)=e+2e+1−log⁡e(e+1)I(1)=e+1e+2​−loge​(e+1).

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