Question

lf $\int \frac{2x^2+3}{(x^2-1)(x^2+4)}dx=a\log \left(\frac{x-1}{x+1}\right)+b{\tan }^{-1}\left(\frac{x}{2}\right)+c$, then values of a and $b$ are

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

Answer 1

The correct option is D $(\frac{1}{2},\frac{1}{2})$

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

$=\frac{1}{2}\log \mid \frac{x-1}{x+1}\mid +\frac{1}{2}{\tan }^{-1}\left(\frac{x}{2}\right)+c$

So, $(a,b)=(\frac{1}{2},\frac{1}{2})$