Question

Let $f\left(x\right)=\frac{1}{e^x+8e^{-x}+4e^{-3x}}$ and $g\left(x\right)=\frac{1}{e^{3x}+8e^x+4e^{-x}}.$ If $\int (f\left(x\right)-2g\left(x\right))dx=h\left(x\right)+C,$ where $C$ is constant of integration and $\underset{x\to \infty }{lim}h\left(x\right)=\frac{\pi }{4},$ then the value of $2\tan \left(2h\right(0\left)\right)$ is

Answer 1

$I=\int (f\left(x\right)-2g\left(x\right))dx$
$=\int \frac{(e^{3x}-2e^x)}{e^{4x}+8e^{2x}+4}dx$
Put $e^x=t\Rightarrow e^{x}dx=dt$
$I=\int \frac{t^2-2}{t^4+8t^2+4}dt$
$=\int \frac{(1-\frac{2}{t^2})}{t^2+8+\frac{4}{t^2}}dt$
$=\int \frac{(1-\frac{2}{t^2})dt}{{(t+\frac{2}{t})}^2+4}$
$=\frac{1}{2}{\tan }^{-1}\left(\frac{t+\frac{2}{t}}{2}\right)+C$
$=\frac{1}{2}{\tan }^{-1}\left(\frac{e^x+2e^{-x}}{2}\right)+C$
$\therefore h\left(x\right)=\frac{1}{2}{\tan }^{-1}\left(\frac{e^x+2e^{-x}}{2}\right)$
$\Rightarrow h\left(0\right)=\frac{1}{2}{\tan }^{-1}\left(\frac{3}{2}\right)$
$\therefore 2\tan \left(2h\right(0\left)\right)=3$