Question

If $\int e^{{\tan }^{-1}\left(\ln x\right)}\left[\frac{2+{\ln }^{2}x}{1+{\ln }^{2}x}\right]dx=g\left(x\right)e^{h\left(x\right)}+C,$ then the value of $h\left(1\right)\cdot g\left(2\right)$ is
(where $C$ is integration constant)

Answer 1

$I=\int e^{{\tan }^{-1}\left(\ln x\right)}\left[\frac{2+{\ln }^{2}x}{1+{\ln }^{2}x}\right]dx=\int e^{{\tan }^{-1}\left(\ln x\right)}\left[\frac{2+{\ln }^{2}x}{1+{\ln }^{2}x}\right]dx=\int e^{{\tan }^{-1}\left(\ln x\right)}[1+\frac{1}{1+{\ln }^{2}x}]dx=\int [e^{{\tan }^{-1}\left(\ln x\right)}+x\cdot \frac{e^{{\tan }^{-1}\left(\ln x\right)}}{x(1+{\ln }^{2}x)}]dx$

Taking $f\left(x\right)=e^{{\tan }^{-1}\left(\ln x\right)}$, then
$I=\int \left[f\right(x)+x{f}^{\prime }(x\left)\right]dx=xf\left(x\right)+C\therefore I=x{e}^{{\tan }^{-1}\left(\ln x\right)}+C\Rightarrow g\left(x\right)=x,h\left(x\right)={\tan }^{-1}\left(\ln x\right)\therefore h\left(1\right)\cdot g\left(2\right)=0$