Question

Evaluate:
$I=\int \frac{x^2-3x+1}{\sqrt{1-x^2}}$

Answer 1

Given,

$\int \frac{x^2-3x+1}{\sqrt{1-x^2}}dx$

$=\int \frac{x^2}{\sqrt{1-x^2}}dx-\int \frac{3x}{\sqrt{1-x^2}}dx+\int \frac{1}{\sqrt{1-x^2}}dx$

$=\frac{1}{2}(\arcsin \left(x\right)-\frac{1}{2}\sin \left(2\arcsin \left(x\right)\right))-(-3\sqrt{1-x^2})+\arcsin \left(x\right)$

$=\frac{3}{2}\arcsin \left(x\right)-\frac{1}{4}\sin \left(2\arcsin \left(x\right)\right)+3\sqrt{-x^2+1}+C$