Question

integrate with respect to x​

Answer 1

Use partial fractions.

u = $sin^{-1}x$

v = 1

∫ uv dx = u∫v dx - ∫ [u'( ∫ vdx)] dx

∫ 1 dx = x + c

u' = 1/$\sqrt{1 - x^{2} }$ dx  

Second term will be:-

∫x/$\sqrt{1-x^{2} }$ dx

1/2∫ 2x / $\sqrt{1-x^{2} }$ dx

Consider 1 - $x^{2}$ = t

-2x dx = dt

So we have

 - $\sqrt{1-x^{2} }$  

First term is:-

$sin^{-1}x$ ∫1dx

Combining the terms as in formula, we have.

x $sin^{-1}x$ - $\sqrt{1-x^{2} }$ +C

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