Question

If $\int \frac{2x^2+3}{(x^2-1)(x^2+4)}dx=Alog\left(\frac{x+1}{x-1}\right)+Bta{n}^{-1}\left(\frac{x}{2}\right)+C$ then (A, B) is

• A. $\left(-\frac{1}{2},\frac{1}{2}\right)$
• B. $\left(\frac{1}{2},-\frac{1}{2}\right)$
• C. $\left(\frac{1}{2},\frac{1}{2}\right)$
• D. $(1,-1)$

Answer 1

The correct option is A $\left(-\frac{1}{2},\frac{1}{2}\right)$

Given, $\int \frac{2x^2+3}{(x^2-1)(x^2+4)}dx$
$=\int (\frac{1}{x^2+4}-\frac{1}{2(x+1)}+\frac{1}{2(x-1)})dx$
$=\int \frac{1}{x^2+4}dx-\frac{1}{2}\int \frac{1}{x+1}dx+\frac{1}{2}\int \frac{1}{x-1}dx$
$=\frac{1}{2}{\tan }^{-1}\left(\frac{x}{2}\right)-\frac{1}{2}\log \left(\frac{x+1}{x-1}\right)$

$\therefore A=-\frac{1}{2},B=\frac{1}{2}$