Question

The value of the integral $\int \frac{x^2}{(x^2+1)(x^2+4)}dx$
(where $C$ is integration constant)

• A. $\frac{1}{3}{\tan }^{-1}x+\frac{2}{3}{\tan }^{-1}\left(\frac{x}{2}\right)+C$
• B. $\frac{1}{3}{\tan }^{-1}x+\frac{2}{3}{\tan }^{-1}\left(\frac{x}{3}\right)+C$
• C. $-\frac{1}{3}{\tan }^{-1}x+\frac{2}{3}{\tan }^{-1}\left(\frac{x}{2}\right)+C$
• D. $-\frac{1}{3}{\tan }^{-1}x+\frac{4}{3}{\tan }^{-1}\left(\frac{x}{2}\right)+C$

Answer 1

The correct option is C $-\frac{1}{3}{\tan }^{-1}x+\frac{2}{3}{\tan }^{-1}\left(\frac{x}{2}\right)+C$

$\int \frac{x^2}{(x^2+1)(x^2+4)}dx=\frac{1}{3}\int [\frac{4}{x^2+4}-\frac{1}{x^2+1}]dx=-\frac{1}{3}\int \frac{1}{x^2+1}dx+\frac{4}{3}\int \frac{1}{x^2+4}dx=-\frac{1}{3}{\tan }^{-1}x+\frac{4}{3}\times \frac{1}{2}{\tan }^{-1}\left(\frac{x}{2}\right)+C=-\frac{1}{3}{\tan }^{-1}x+\frac{2}{3}{\tan }^{-1}\left(\frac{x}{2}\right)+C$