Definite Integration
Indefinite Integration
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: Use substitution u = ln((x-1)/(x+1)). Then du = (1/((x-1)/(x+1))) * d/dx((x-1)/(x+1)) dx = ((x+1)/(x-1)) * ((x+1 - (x-1))/(x+1)^2) dx = ((x+1)/(x-1)) * (2/(x+1)^2) dx = 2/(x^2-1) dx. Thus, the integral becomes 1/2 * integral(u du) = u^2/4 + C.
Let $I = \int \frac{\ln\left(\frac{x-1}{x+1}\right)}{x^2-1} dx$. Let $u = \ln\left(\frac{x-1}{x+1}\right)$. Then $du = \frac{x+1}{x-1} \cdot \frac{(x+1) - (x-1)}{(x+1)^2} dx = \frac{2}{(x-1)(x+1)} dx = \frac{2}{x^2-1} dx$. So $\frac{dx}{x^2-1} = \frac{du}{2}$. Thus $I = \int u \cdot \frac{du}{2} = \frac{u^2}{4} + C = \frac{1}{4} \ln^2 \left(\frac{x-1}{x+1}\right) + C$. Since $\ln^2 \left(\frac{x-1}{x+1}\right) = \ln^2 \left(\frac{x+1}{x-1}\right)$, both (B) and (D) are correct.
Correct Answer: B,D