$\int \frac{x\cos x+1}{\sqrt{2x^3{e}^{\sin x}+x^2}}dx$

Solution:

$I=\int \frac{x\cos x+1}{\sqrt{2x^3{e}^{\sin x}+x^2}}dx$

$=\int \frac{x\cos x+1}{x\sqrt{2x{e}^{\sin x}+1}}dx$

Put $2x{e}^{\sin x}+1=t$

$\Rightarrow (2x\left(e^{\sin x}\right)\cos x+e^{\sin x}\left(2\right))dx=dt$

$\Rightarrow 2e^{\sin x}[x\cos x+1]dx=dt$

$\Rightarrow 2x{e}^{\sin x}\left(\frac{{\displaystyle x\cos x+1}}{{\displaystyle x}}\right)dx=dt$

$\Rightarrow (t-1)\left(\frac{{\displaystyle x\cos x+1}}{{\displaystyle x}}\right)dx=dt$

$\Rightarrow \left(\frac{{\displaystyle x\cos x+1}}{{\displaystyle x}}\right)dx=\frac{dt}{(t-1)}$

$Now I=\int \frac{1}{(t-1)\sqrt{t}}dt$

$t=z^2\Rightarrow dt=2zdz$

$I=\int \frac{2z}{(z^2-1)\left(z\right)}dz$

$=2\left(\frac{1}{2}\right)\ln \left|\frac{z-1}{z+1}\right|+c$

$where z=\sqrt{t}=\sqrt{2x{e}^{\sin x}+1}$

$I=\ln \left|\frac{{\displaystyle \sqrt{2x{e}^{\sin x}+1}-1}}{{\displaystyle \sqrt{2x{e}^{\sin x}+1}+1}}\right|+c$