Question
Integrate the following w.r.t. x
$\frac{x^{2} - 3x + 1}{\sqrt{1 - x^{2}}}$

Answer

$\int\frac{\text{x}^{2} - 3\text{x} + 1}{\sqrt{1 - \text{x}^{2}}} \text{dx} = \int\frac{2 - 3\text{x} - (1 - \text{x}^{2})}{\sqrt{1 - \text{x}^{2}}} \text{dx}$
$= 2\int\frac{1}{\sqrt{1 - \text{x}^{2}}} \text{dx} -3 \int\frac{\text{x}}{\sqrt{1 - \text{x}^{2}}} \text{dx} - \int\sqrt{1 - \text{x}^{2}} \text{dx}$
$= 2 \sin^{-1}\text{x} + 3 \sqrt{1 - \text{x}^{2}} - \frac{\text{x}}{2} \sqrt{1 - {\text{x}^{2}}} - \frac{1}{2} \sin^{-1} \text{x + c}$
$\text{or} = \frac{3}{2} \sin^{-1} \text{x} + \frac{1}{2} (6 - \text{x}) \sqrt{1 - \text{x}^{2}} + \text{c}$

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