Definite Integration
Substitution method
Grade Class 12
Question:
The value of $\int \frac{\ln\left(\frac{x-1}{x+1}\right)}{x^2 - 1} dx$ is equal to
$\frac{1}{2} \ln^2 \frac{x-1}{x+1} + C$
$\frac{1}{4} \ln^2 \frac{x-1}{x+1} + C$
$\frac{1}{2} \ln^2 \frac{x+1}{x-1} + C$
$\frac{1}{4} \ln^2 \frac{x+1}{x-1} + C$
Step-by-Step Solution
Key Concept: Substitution method
Let t = ln((x-1)/(x+1)). Then dt = (2/(x^2-1)) dx.
Correct Answer: B,D