Question

If $\int \frac{dx}{(x^2-9)\sqrt{x+1}}=p\cdot \ln \left|\frac{f\left(x\right)-2}{f\left(x\right)+2}\right|+q\cdot {\tan }^{-1}\left(g\right(x\left)\right)+C,$ then the value of $p\cdot f\left(3\right)+q\cdot g\left(0\right)$ is
(Where $p,q$ are fixed constants and $C$ is integration constant)

Answer 1

Let $I=\int \frac{dx}{(x^2-9)\sqrt{x+1}}$
Substituting $x+1=t^2,dx=2tdt$
Integral becomes,
$I=\int \frac{2tdt}{\left(\right(t^2-1{)}^2-9)t}=\int \frac{2dt}{(t^2-4)(t^2+2)}=\frac{1}{3}\int [\frac{1}{t^2-4}-\frac{1}{t^2+2}]dt=\frac{1}{12}\ln \left|\frac{t-2}{t+2}\right|-\frac{1}{3\sqrt{2}}{\tan }^{-1}\left(\frac{t}{\sqrt{2}}\right)+C=\frac{1}{12}\ln \left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right|-\frac{1}{3\sqrt{2}}{\tan }^{-1}\left(\sqrt{\frac{x+1}{2}}\right)+C$
$\Rightarrow p=\frac{1}{12},q=-\frac{1}{3\sqrt{2}},f\left(x\right)=\sqrt{x+1},g\left(x\right)=\sqrt{\frac{x+1}{2}}$
So, $p\cdot f\left(3\right)+q\cdot g\left(0\right)=0$