antiderivitive of x(1+x^2)^1/2

[Guest]

I need to find:

. . .x(1+x^2)^1/2

Or:

. . .x / (1 + x^2)^(1/2)

I need the anti derivitive of either one, which ever is easier. (They are both the same for my purposes.) I got stuck in the middle of:

. . .integral {from 0 to 1/sqrt[2]} arcsin xdx

thanks alot

 

[soroban]

Hello, causalitist!

\(\displaystyle \L\int^{\;\;\;\frac{1}{\sqrt{2}}}_0 \arcsin x\,dx\)

Integrate by parts:

Let: \(\displaystyle \,u\,=\,\arcsin x\;\;\;\;\;\;dv\,=\,dx\)

Then: \(\displaystyle \,du\,=\,\frac{dx}{\sqrt{1\,-\,x^2}}\;\;\;\;\;v\,=\,x\)

We have: \(\displaystyle \L\,x\cdot\arcsin x \,-\,\int x(1\,-\,x^2)^{-\frac{1}{2}}dx\)

. . . \(\displaystyle \L= \;x\cdot\arcsin x \,-\,\sqrt{1\,-\,x^2}\,\bigg]^{\frac{1}{\sqrt{2}}}_0\)

. . . \(\displaystyle \L= \;\bigg[\frac{1}{\sqrt{2}}\arcsin\left(\frac{1}{\sqrt{2}}\right)\,-\,\sqrt{1\,-\,\left(\frac{1}{\sqrt{2}}\right)^2}\bigg ] \,- \,\bigg[0\cdot\arcsin 0 \,-\,\sqrt{1\,-\,0^2}\bigg]\)

. . . \(\displaystyle \L=\;\frac{1}{\sqrt{2}}\cdot\frac{\pi}{4} \,-\,\frac{1}{\sqrt{2}}\,+\,1 \;= \;\frac{\pi\,-\,4\,+\,4\sqrt{2}}{4\sqrt{2}}\)