# 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?

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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 $$I=2\arcsin (x)\sqrt{1+x}-\int \frac{2}{\sqrt{1-x}}\,\mathrm dx=2\arcsin (x)\sqrt{1+x}+4\sqrt{1-x}.$$
Integrate by parts, letting $dv = (x+1)^{-1/2}$ and $u = \arcsin(x)$, then $du = \frac{1}{\sqrt{1-x^2}}$ and $v = \int (x+1)^{-1/2} dx = 2 \sqrt{x+1}$.
Use $\int u dv = uv - \int v du$ to reduce this to
$$\int v du = \int 2 \sqrt{ \frac{x+1}{1-x^2} } dx = 2 \int \frac{dx}{\sqrt{1-x}} = -4 \sqrt{1-x}.$$