$\int (\sin^{-1} x + \cos^{-1} x) \, dx = $

$\int \frac{x+\sin x}{1+\cos x} d x=$

Let $g:(0, \infty) \rightarrow R$ be a differentiable function such that $\int \left( \frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - x e^x)}{(e^x + 1)^2} \right) dx = \frac{x g(x)}{e^x + 1} + c$ for all $x > 0$,where $c$ is an arbitrary constant. Then:

If $\int \frac{\cos x+x}{1+\sin x} d x=f(x)+\int \frac{3 \cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} d x+c_r$ then $f(x)=$

Let $x > 0$ be a fixed real number. Then,the integral $\int \limits_0^{\infty} e^{-t}|x-t| d t$ is equal to

$\int e^{2 x}\left[\cos (3 x+4)+5 x^2\right] d x=$