Question

If $\int \frac{1}{(x^2+4)(x^2+9)}dx=A{\tan }^{-1}\frac{x}{2}+B{\tan }^{-1}\left(\frac{x}{3}\right)+C$ then $A-B=$

• A. $\frac{1}{6}$
• B. $\frac{1}{30}$
• C. $-\frac{1}{30}$
• D. $-\frac{1}{6}$

Answer 1

The correct option is A $\frac{1}{6}$

Given : $\therefore \int \frac{1}{(x^2+4)(x^2+9)}dx=A{\tan }^{-1}\frac{x}{2}+B{\tan }^{-1}\frac{x}{3}+C$ ...... $\left(i\right)$

$\frac{1}{AB}=\frac{1}{B-A}(\frac{1}{A}-\frac{1}{B})$

$\therefore \int \frac{1}{(x^2+4)(x^2+9)}dx=\int \frac{1}{5}(\frac{1}{x^2+4}-\frac{1}{x^2+9})dx$

$=\frac{1}{5}[\frac{1}{2}{\tan }^{-1}\frac{x}{2}-\frac{1}{3}{\tan }^{-1}\frac{x}{3}]+C$

$=\frac{1}{10}{\tan }^{-1}\frac{x}{2}-\frac{1}{15}{\tan }^{-1}\frac{x}{3}+C$

Comparing above equation with $\left(i\right)$ we get

$A=\frac{1}{10}$ and $B=-\frac{1}{15}$

$\therefore A-B=\frac{1}{10}+\frac{1}{15}=\frac{5}{30}=\frac{1}{6}$