How do I evaluate $int (arcsin x)/sqrt (1+x) dx$ ?

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Answer 1 · 2015-02-17T22:53:26

We will integrate this using integration by parts.

Recall that $uv-intvdu$

Let $u=arcsin(x)$ and $dv=dx/sqrt(1+x)$

Differentiating $u=arcsinx$ we have
$du=1/sqrt(1-x^2)dx$

Integrate $dv=dx/sqrt(1+x)$

$intdv=intdx/sqrt(1+x)$

$v=2sqrt(1+x)$

Therefore,

$arcsin(x)(2sqrt(1+x))-int2sqrt(1+x)(1/sqrt(1-x^2))dx$

Some rewriting and factor $1-x^2$

$2arcsin(x)sqrt(1+x)- 2intsqrt(1+x)/sqrt((1+x)(1-x))dx$

$2arcsin(x)sqrt(1+x)-2intsqrt(1+x)/(sqrt(1+x)sqrt(1-x))dx$

$2arcsin(x)sqrt(1+x)-2int1/sqrt(1-x)dx$

Finally, we integrate

$2arcsin(x)sqrt(1+x)+4sqrt(1-x)+C$

How do I evaluate int (arcsin x)/sqrt (1+x) dxint (arcsin x)/sqrt (1+x) dx?