Question

Evaluate: \[ \int \frac{x+1}{x(1+xe^x)^2}\,dx \]

• (A) \[ \ln\left|\frac{xe^x}{1+xe^x}\right|+\frac{1}{1+xe^x}+C \]
• (B) \[ \ln|xe^x|-\ln|1+xe^x|-\frac{1}{1+xe^x}+C \]
• (C) \[ \ln\left|\frac{1+xe^x}{xe^x}\right|+\frac{1}{1+xe^x}+C \]
• (D) \[ \frac{1}{1+xe^x}+C \]

Hint:

Whenever the expression \( xe^x \) appears repeatedly inside brackets and denominators, always check whether its derivative \( e^x(x+1) \) is present in the numerator.

Answer 1

The Correct Option is A

Concept:
Whenever expressions involving \( xe^x \) appear in integrals, substitution with \[ t=xe^x \] is generally very effective because: \[ \frac{d}{dx}(xe^x)=e^x(x+1). \]

Step 1: Preparing the integrand for substitution.
Given, \[ I=\int \frac{x+1}{x(1+xe^x)^2}\,dx. \] Multiply numerator and denominator by \( e^x \): \[ I=\int \frac{(x+1)e^x}{xe^x(1+xe^x)^2}\,dx. \] Now let \[ t=xe^x. \] Then, \[ dt=e^x(x+1)\,dx. \] Hence, \[ I=\int \frac{dt}{t(1+t)^2}. \]

Step 2: Using partial fractions.
Resolve: \[ \frac{1}{t(1+t)^2} = \frac{A}{t} +\frac{B}{1+t} +\frac{C}{(1+t)^2}. \] Multiplying throughout by \( t(1+t)^2 \): \[ 1=A(1+t)^2+Bt(1+t)+Ct. \] Expanding: \[ 1=A(1+2t+t^2)+Bt+Bt^2+Ct. \] Grouping like terms: \[ 1=A+(2A+B+C)t+(A+B)t^2. \] Comparing coefficients: \[ A=1, \] \[ A+B=0, \] \[ 2A+B+C=0. \] Substituting \( A=1 \): \[ 1+B=0 \Rightarrow B=-1. \] Again, \[ 2(1)-1+C=0 \Rightarrow 1+C=0 \Rightarrow C=-1. \] Therefore, \[ \frac{1}{t(1+t)^2} = \frac1t-\frac1{1+t}-\frac1{(1+t)^2}. \]

Step 3: Integrating term by term.
Thus, \[ I= \int\left( \frac1t-\frac1{1+t}-\frac1{(1+t)^2} \right)dt. \] Integrating: \[ I=\ln|t|-\ln|1+t|+\frac1{1+t}+C. \]

Step 4: Substituting back \( t=xe^x \).
Hence, \[ I= \ln\left| \frac{xe^x}{1+xe^x} \right| + \frac1{1+xe^x} +C. \] Therefore, \[ \boxed{ \ln\left| \frac{xe^x}{1+xe^x} \right| + \frac1{1+xe^x} +C }. \]