Definite Integral (4)

\int_{0}^{\frac{\pi}{2}}{\frac{\left(\sqrt{1+x^2} \right)\cos\left( {x+\arctan{x}\right)\text{dx}}}{x^2+\cos^2x}}

may be we can do this

$ Original Question is $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{cosx+xsinx}{x^2+cos^2x}dx\\\\ $Sorry for my blunder typing........

then it is easy , it is

\tan^{-1}{\left( x\sec x\right)}

\int \frac{\left( \cos x +x \sin x\right)\text{dx}}{\cos^2 x\left(1+\left( \frac{x}{\cos x}\right) ^2\right)}
\text{Now put }\frac{x}{\cos x}=t

\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{cosx-xsinx}{x^2+cos^2x}dx = \pi$