Question

$\int \frac{dx}{(x^2+1)(x^2+4)}=$

• A. $\frac{1}{3}{\tan }^{-1}x-\frac{1}{3}{\tan }^{-1}\frac{x}{2}+c$
• B. $\frac{1}{3}{\tan }^{-1}x+\frac{1}{3}{\tan }^{-1}\frac{x}{2}+c$
• C. $\frac{1}{3}{\tan }^{-1}x-\frac{1}{6}{\tan }^{-1}\frac{x}{2}+c$
• D. ${\tan }^{-1}x-2{\tan }^{-1}\frac{x}{2}+c$

Answer 1

The correct option is C $\frac{1}{3}{\tan }^{-1}x-\frac{1}{6}{\tan }^{-1}\frac{x}{2}+c$

Given,

$\int \frac{dx}{(x^2+1)(x^2+4)}$

$=\frac{1}{3}\int \frac{3dx}{(x^2+1)(x^2+4)}$

$=\frac{1}{3}\int \frac{x^2+4-(x^2+1)}{(x^2+1)(x^2+4)}dx$

$=\int \frac{1}{3(x^2+1)}dx-\frac{1}{3(x^2+4)}dx$

$=\frac{1}{3}{\tan }^{-1}\left(x\right)-\frac{1}{6}{\tan }^{-1}\left(\frac{x}{2}\right)+C$