Question

The value of the integral $\int \frac{1}{x^4-1}dx$ is
(where $C$ is an arbitrary constant)

• A. $\frac{1}{4}\ln \left|\frac{x-1}{x+1}\right|+\frac{1}{2}{\tan }^{-1}\frac{x}{2}+C$
• B. $\frac{1}{2}\ln \left|\frac{x-1}{x+1}\right|-\frac{1}{2}{\tan }^{-1}x+C$
• C. $\frac{1}{2}\ln \left|\frac{x-1}{x+1}\right|+\frac{1}{4}{\tan }^{-1}\frac{x}{2}+C$
• D. $\frac{1}{4}\ln \left|\frac{x-1}{x+1}\right|-\frac{1}{2}{\tan }^{-1}x+C$

Answer 1

The correct option is D $\frac{1}{4}\ln \left|\frac{x-1}{x+1}\right|-\frac{1}{2}{\tan }^{-1}x+C$

$\int \frac{1}{(x^4-1)}dx=\int \frac{1}{(x^2-1)(x^2+1)}dx=\int \frac{1}{2}(\frac{1}{x^2-1}-\frac{1}{x^2+1})dx=\int \frac{1}{2(x^2-1)}dx-\int \frac{1}{2(x^2+1)}dx=\frac{1}{4}\ln \left|\frac{x-1}{x+1}\right|-\frac{1}{2}{\tan }^{-1}x+C$