Indefinite IntegrationhardFree

Find f(x) from ∫(x+1)/(x(1+xeˣ)²)dx | JEE

JEE Maths question with a full step-by-step solution.

Question

\displaystyle\int \dfrac{x+1}{x(1+xe^{x})^{2}}\,dx=\log|1-f(x)|+f(x)+C

, then

f(x)

equals

\dfrac{1}{x+e^{x}}

\dfrac{1}{1+xe^{x}}

correct

\dfrac{1}{(1+xe^{x})^{2}}

\dfrac{1}{(1+e^{x})^{2}}

Solution

Step 1: The factor

(1+xe^{x})

suggests the substitution

t=xe^{x}

. Then

dt=(e^{x}+xe^{x})\,dx=(x+1)e^{x}\,dx

Step 2: Introduce

e^{x}

to match

dt

. Multiply numerator and denominator by

e^{x}

I=\int\dfrac{(x+1)e^{x}}{xe^{x}(1+xe^{x})^{2}}\,dx=\int\dfrac{dt}{t(1+t)^{2}}

Step 3: Resolve into partial fractions. Writing

\dfrac{1}{t(1+t)^{2}}=\dfrac{A}{t}+\dfrac{B}{1+t}+\dfrac{D}{(1+t)^{2}}

gives

A=1,\ B=-1,\ D=-1

\dfrac{1}{t(1+t)^{2}}=\dfrac{1}{t}-\dfrac{1}{1+t}-\dfrac{1}{(1+t)^{2}}

Step 4: Integrate each term:

I=\ln|t|-\ln|1+t|+\dfrac{1}{1+t}+C=\ln\left|\dfrac{t}{1+t}\right|+\dfrac{1}{1+t}+C

Step 5: Match the required form. Since

\dfrac{t}{1+t}=1-\dfrac{1}{1+t}

I=\log\left|1-\dfrac{1}{1+t}\right|+\dfrac{1}{1+t}+C

Comparing with

\log|1-f(x)|+f(x)+C

gives

f(x)=\dfrac{1}{1+t}=\dfrac{1}{1+xe^{x}}

. Correct answer: (2)

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