Find x^2-3 x+1 1-x^2 d x

Question

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

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Answer

Let $I=\int \frac{x^2-3 x+1}{\sqrt{1-x^2}}$
$=(-1) \int \frac{-x^2+3 x-1}{\sqrt{1-x^2}} d x$
$=(-1) \int \frac{-x^2+3 x-1+1-1}{\sqrt{1-x^2}} d x$
$=(-1) \int \frac{1-x^2+3 x-2}{\sqrt{1-x^2}} d x$
$=(-1) \int\left[\frac{1-x^2}{\sqrt{1-x^2}}+\frac{3 x-2}{\sqrt{1-x^2}}\right] d x$
$=(-1) \int\left[\sqrt{1-x^2}+\int \frac{3 x-2}{\sqrt{1-x^2}}\right] d x$
$=(-1)\left[\int \sqrt{1-x^2} d x+\int \frac{3 x-2}{\sqrt{1-x^2}} d x\right]$
$=(-1)\left(I_1+I_2\right) \ldots \ldots \text { (i) }$
$\text { consider, } I_1=\int \sqrt{1-x^2} d x$
$=\frac{1}{2}\left[x \sqrt{1-x^2}+\sin ^{-1}(x)\right]+C_1 \ldots \text { (ii) }\left[\because \int \sqrt{a^2-x^2} d x=\frac{1}{2}\left[x \sqrt{a^2-x^2}+a^2 \sin ^{-1}\left(\frac{x}{a}\right)\right]+C\right)$
$\text { consider } I_2=\int \frac{3 x-2}{\sqrt{1-x^2}} d x$
$=\int \frac{3 x}{\sqrt{1-x^2}} d x-2 \int \frac{d x}{\sqrt{1-x^2}}$
$=-\frac{3}{2} \int \frac{-2 x}{\sqrt{1-x^2}} d x-2 \int \frac{d x}{\sqrt{1-x^2}}$
$=-\frac{3}{2} \times 2 \sqrt{1-x^2}-2 \sin ^{-1}(x)+C_2\left[\because \int \frac{f^{\prime}(x)}{\sqrt{f(x)}} d x=2 \sqrt{f(x)}+C\right]$
$=-3 \sqrt{1-x^2}-2 \sin ^{-1}(x)+C_2 \ldots \text { (iii) }$
From Equations $(i), (ii)$ and $(iii),$ we get
$I=(-1)\left(\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1}(x)+C_1-2 \sin ^{-1}(x)-3 \sqrt{1-x^2}+C_2\right)$
$I=\frac{3}{2} \sin ^{-1}(x)-\frac{x}{2} \sqrt{1-x^2}+3 \sqrt{1-x^2}+C\ ($where $C=-C_1-C_2)$