Question:medium

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

Show 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.
Updated On: May 29, 2026
  • (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) \[ \frac{1}{1+xe^x}+C \]
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
This is an indefinite integration problem involving transcendental functions.
The integrand contains a mixture of polynomials and exponentials, specifically the term \( x e^x \).
A common technique for such integrals is to find a substitution that simplifies the expression into a rational function.
We notice that the derivative of \( x e^x \), using the product rule, is \( e^x(x + 1) \).
The numerator contains \( x+1 \). If we can introduce \( e^x \) into the numerator, we can use the substitution \( t = x e^x \).
Step 2: Key Formula or Approach:
1. Multiply numerator and denominator by \( e^x \).
2. Let \( t = x e^x \), then \( dt = (x+1)e^x dx \).
3. Use partial fraction decomposition for the resulting integral \( \int \frac{dt}{t(1+t)^2} \).
Step 3: Detailed Explanation:
Let the integral be \( I \):
\[ I = \int \frac{x + 1}{x(1 + x e^x)^2} dx \]
Multiply both top and bottom by \( e^x \):
\[ I = \int \frac{(x + 1)e^x}{x e^x(1 + x e^x)^2} dx \]
Now, put \( t = x e^x \). Differentiating:
\[ dt = (x \cdot e^x + e^x \cdot 1) dx = e^x(x+1) dx \]
The integral becomes:
\[ I = \int \frac{dt}{t(1 + t)^2} \]
We decompose the fraction using partial fractions:
\[ \frac{1}{t(1 + t)^2} = \frac{A}{t} + \frac{B}{1 + t} + \frac{C}{(1 + t)^2} \]
Multiply by \( t(1 + t)^2 \):
\[ 1 = A(1 + t)^2 + Bt(1 + t) + Ct \]
- If \( t = 0 \): \( 1 = A(1)^2 \implies A = 1 \).
- If \( t = -1 \): \( 1 = C(-1) \implies C = -1 \).
- Compare coefficients of \( t^2 \): \( 0 = A + B \implies B = -A = -1 \).
The partial fraction decomposition is:
\[ \frac{1}{t(1 + t)^2} = \frac{1}{t} - \frac{1}{1 + t} - \frac{1}{(1 + t)^2} \]
Integrating this:
\[ I = \int \left( \frac{1}{t} - \frac{1}{1 + t} - \frac{1}{(1 + t)^2} \right) dt \]
\[ I = \ln|t| - \ln|1 + t| - \left( -\frac{1}{1 + t} \right) + C \]
\[ I = \ln\left| \frac{t}{1 + t} \right| + \frac{1}{1 + t} + C \]
Substitute \( t = x e^x \) back into the expression:
\[ I = \ln\left| \frac{x e^x}{1 + x e^x} \right| + \frac{1}{1 + x e^x} + C \]
This matches Option (A).
Step 4: Final Answer:
Through a strategic substitution and partial fraction decomposition, we converted the transcendental integral into a form involving natural logarithms and a simple rational term. This corresponds to Option (A).
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