# Part of integration

Hello. I was solving $\int x\arcsin x\, dx$. When applying integration by parts, the book suggests ($v=2x^2-1$), so I came up with this:
$$\displaystyle \frac{1}{4}\left[(2x^2-1)\arcsin x - \!\int \frac {2x^2-1}{\sqrt{1-x^2}}dx\right]$$. I don't have any idea how the second part is solved.
I mean this part: $$\displaystyle (- \!\int \frac {2x^2-1}{\sqrt{1-x^2}}dx)$$.
Thank you all.

#### idontknow

$$\displaystyle \dfrac{2x^2 -1}{\sqrt{1-x^2 }}dx=\dfrac{2x^2 -2}{\sqrt{1-x^2 }}dx +\dfrac{dx}{\sqrt{1-x^2 }}=-2\sqrt{1-x^2 }dx +\dfrac{dx}{\sqrt{1-x^2 }}.$$

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#### skipjack

Forum Staff
$$\displaystyle -\!\int\frac{2x^2 - 1}{\sqrt{1 - x^2}}\,dx = \!\int\left(\frac{1 - x^2}{\sqrt{1 - x^2}} - \frac{x^2}{\sqrt{1 - x^2}}\right)dx = \!\int \left(\sqrt{1 - x^2} + x\cdot\frac{-(2x)/2}{\sqrt{1 - x^2}}\right)dx$$

etc.