Question

integral of (arccos(x))/(sqrt 1+x)

Answer 1

you may proceed by assumin
x=cos u
then dx = -sinu du = -2*sin(u/2)*cos(u/2) du
and 1+x =sqrt(2) * cos (u/2)

u may use integration by parts here

Answer 2

I think you can just use integration by parts from the start too:
let
\[u = \arccos x \implies du=-\frac{1}{\sqrt{1-x^2}}dx\\
dv = \frac{1}{\sqrt{1+x}}dx\implies v=2\sqrt{x+1}\]
Then, \[ \int\frac{\arccos x}{\sqrt{1+x}}dx=\arccos x\cdot2\sqrt{x+1}-\int2\sqrt{x+1}\left( -\frac{1}{\sqrt{1-x^2}}\right)\,dx\\
\, \\
= 2\sqrt{x+1}\,(\arccos x)+2\int \frac{\sqrt{1+x}}{\sqrt{(1-x)(1+x)}}\,dx \\
\, \\
=2\sqrt{x+1}\,(\arccos x)+2\int \frac{1}{\sqrt{1-x}}\,dx\]. The rest should be easy with simple u-substitution.