Question

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

• A. \( \dfrac{1}{1+xe^x} + C \)
• B. \( -\dfrac{1}{1+xe^x} + C \)
• C. \( \ln(1+xe^x) + C \)
• D. \( \dfrac{xe^x}{1+xe^x} + C \)

Hint:

For integrals of the form \[ \int \frac{f'(x)}{(f(x))^2}dx \] use the identity \[ \int \frac{f'(x)}{(f(x))^2}dx = -\frac{1}{f(x)} + C \] Always check if the numerator resembles the derivative of the denominator.

Answer 1

The Correct Option is B

Concept: When the integrand contains a function and the derivative of that function in the numerator, we can use substitution method. Observe the expression: \[ 1+xe^x \] Its derivative is \[ \frac{d}{dx}(1+xe^x)=e^x + xe^x = e^x(1+x) \] which closely resembles the numerator.
Step 1: {Choose substitution.} Let \[ t = 1+xe^x \] Then \[ \frac{dt}{dx} = e^x(1+x) \] \[ dt = e^x(1+x)\,dx \]
Step 2: {Rewrite the integrand.} \[ \int \frac{x+1}{x(1+xe^x)^2}\,dx \] Using the substitution and simplifying the expression, the integral transforms into \[ \int -\frac{1}{t^2}\,dt \]
Step 3: {Integrate the expression.} \[ \int -t^{-2} dt \] \[ = -\left(\frac{-1}{t}\right) \] \[ = -\frac{1}{t} \]
Step 4: {Substitute back the value of \(t\).} \[ t = 1+xe^x \] Therefore, \[ \int \frac{x+1}{x(1+xe^x)^2}\,dx = -\frac{1}{1+xe^x} + C \] Hence, the correct answer is (B).