Evaluate $ \int \frac{\sin^{-1}(x)}{\sqrt{1+x}} \; dx$ So I have to find the integral of $$ \int  \frac{\sin^{-1}(x)}{\sqrt{1+x}} \; dx$$
I think I have to do this using the integration by parts..so I will take $f = \sin^{-1}(x)$ and $ \sqrt {1+x}=g' $...what about now? 

Do integration by parts with $f(x)=\arcsin(x)$ and $g(x)=\frac{1}{\sqrt{1+x}}$. Then
$$
I=\int\frac{\arcsin x}{\sqrt{1+x}}\,\mathrm dx=\int f(x)g(x)\,\mathrm dx=f(x)G(x)-\int f'(x)G(x)\,\mathrm dx\\
=2\arcsin (x)\sqrt{1+x}-\int \frac{1}{\sqrt{1-x^2}}\cdot2\sqrt{1+x}\,\mathrm dx,
$$
but
$$
\frac{1}{\sqrt{1-x^2}}\cdot2\sqrt{1+x}=\frac{2\sqrt{1+x}}{\sqrt{1+x}\sqrt{1-x}}=\frac{2}{\sqrt{1-x}},
$$
so
$$
I=2\arcsin (x)\sqrt{1+x}-\int \frac{2}{\sqrt{1-x}}\,\mathrm dx=2\arcsin (x)\sqrt{1+x}+4\sqrt{1-x}.
$$
