Question

pl solve attached maths integral

Answer 1

$\int\frac{x\tan^{-1}x}{\sqrt{1+x^2}}$

substitute $t=\tan^{-1}(x)\implies dt=\frac{dx}{1+x^2}$ and we get
$=\int t\sec(t)\tan(t)\,dt$

Next integrate by parts letting
$t=u$ and $dv=\sec(t)\tan(t)$,
$dt=du$ and $v=\sec(t)$ : 

$=t\sec(t)-\int \sec(t)\,dt$
$=t\sec(t)-\ln|\sec(t)+\tan(t)|+C$

Plugging back $t=\tan^{-1}(x)$ gives
$=\tan^{-1}(x)\sqrt{1+x^2}-\ln|\sqrt{1+x^2}+x|+C$