Integrating $\int \frac{\sin^{-1}(x)}{\sqrt{1+x}}dx$ by parts

Question

I have a question that requires me to integrate the following by parts. I have done the question but apparently my answer does not match that of wolfram alpha's.

$$\int \frac{\sin^{-1}(x)}{\sqrt{1+x}}dx$$ $$ u= \sin^{-1}(x) : du = \frac{1}{\sqrt{1-x^2}}dx$$ $$ v= 2\sqrt{1+x} : dv = \frac{1}{\sqrt{1+x}}dx$$ Following the formula $\int{udv}=uv-\int{vdu}$ $$=2\sqrt{1+x}*(\sin^{-1}(x))-\int{\frac{\sqrt{1+x}}{\sqrt{(1+x)(1-x)}}dx}$$ $$=2\sqrt{1+x}*(\sin^{-1}(x))-\int{\frac1{\sqrt{1-x}}}dx$$ $$=2\sqrt{1+x}*(\sin^{-1}(x))-4\sqrt{1-x} +C$$

But contrary to my answer... the wolfram alpha answer provided is...

$$\frac{2[2\sqrt{1-x^2}+(x+1)(\sin^{-1}(x))]}{\sqrt{x+1}}$$

What am I doing wrong, or ... how do I simplify to that form?

Answer 1

First, you wrote

$$-\int\frac{\sqrt{1+x}}{\sqrt{(1+x)(1-x)}}dx\tag 1$$

where $(1)$ should have been

$$-2\int\frac{\sqrt{1+x}}{\sqrt{(1+x)(1-x)}}dx \tag 2$$

This was likely a typographical error (i.e., omitting the $2$). This error carried to the next line in which you wrote

$$-\int\frac{1}{\sqrt{1-x}}dx$$

where it should have been

$$-2\int\frac{1}{\sqrt{1-x}}dx$$

But, the next error is, I believe, the source of the issue. You integrated $-\int \frac{1}{\sqrt{1-x}}dx$ and obtained

$$-4\sqrt{1-x}$$

Surely,

$$-\int \frac{1}{\sqrt{1-x}}dx=+2\sqrt{1-x}$$

If we insert the missing factor of $2$ as in $(2)$, we have

$$-2\int\frac{\sqrt{1+x}}{\sqrt{(1+x)(1-x)}}dx =+4\sqrt{1-x}+C$$

and the final answer after correction is

$$\int\frac{\arcsin x}{\sqrt{1+x}}dx=2\sqrt{1+x}\arcsin x+4\sqrt{1-x}+C$$

which agrees with WA after multiplying by $1=\frac{\sqrt{1+x}}{\sqrt{1+x}}$!

Answer 2

$$\frac{2[2\sqrt{1-x^2}+(x+1)(\sin^{-1}(x))]}{\sqrt{x+1}}=\frac{4\overbrace{\sqrt{1-x^2}}^{=\sqrt{1+x}\sqrt{1-x}}}{\sqrt{x+1}}+2\frac{x+1}{\sqrt{x+1}}\arcsin x=$$

$$4\sqrt{1-x}+2\sqrt{1+x}\arcsin x$$

Also

$$\int\frac{dx}{\sqrt{1-x}}=\color{red}-2\sqrt{1-x}+C$$