Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{x}{\sqrt{8+x-x^{2}}} d x$

Solution:

Given $I=\int \frac{x}{\sqrt{-x^{2}+x+8}} d x$

Integral is of form $\int \frac{p x+q}{\sqrt{a x^{2}+b x+c}} d x$

Writing numerator as $\mathrm{px}+\mathrm{q}=\lambda\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)\right\}+\mu$

$\Rightarrow p x+q=\lambda(2 a x+b)+\mu$

$\Rightarrow x=\lambda(-2 x+1)+\mu$

$\therefore \lambda=-1 / 2$ and $\mu=-1 / 2$

Let $x=-1 / 2(-2 x+1)-1 / 2$ and split,

$\Rightarrow \int \frac{x}{\sqrt{-x^{2}+x+8}} d x=\int\left(\frac{-(-2 x+1)}{2 \sqrt{-x^{2}+x+8}}-\frac{1}{2 \sqrt{-x^{2}+x+8}}\right) d x$

$=\frac{1}{2} \int \frac{2 x-1}{\sqrt{-x^{2}+x+8}} d x-\frac{1}{2} \int \frac{1}{\sqrt{-x^{2}+x+8}} d x$

Consider $\int \frac{2 x-1}{\sqrt{-x^{2}+x+8}} d x$

Let $u=-x^{2}+x+8 \rightarrow d x=\frac{1}{-2 x+1} d u$

$\Rightarrow \int \frac{2 \mathrm{x}-1}{\sqrt{-\mathrm{x}^{2}+\mathrm{x}+8}} \mathrm{dx}=\int-\frac{1}{\sqrt{\mathrm{u}}} \mathrm{du}$

$=-\int \frac{1}{\sqrt{\mathrm{u}}} \mathrm{du}$

We know that $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$

$\Rightarrow-\int \frac{1}{\sqrt{u}} d u=-(2 \sqrt{u})$

$=-2 \sqrt{-x^{2}+x+8}$

Consider $\int \frac{1}{\sqrt{-x^{2}+x+8}} d x$

$\Rightarrow \int \frac{1}{\sqrt{-x^{2}+x+8}} d x=\int \frac{1}{\sqrt{\frac{33}{4}-\left(x-\frac{1}{2}\right)^{2}}} d x$

Let $\mathrm{u}=\frac{2 \mathrm{x}-1}{\sqrt{33}} \rightarrow \mathrm{dx}=\frac{\sqrt{33}}{2} \mathrm{du}$

$\Rightarrow \int \frac{1}{\sqrt{\frac{33}{4}-\left(x-\frac{1}{2}\right)^{2}}} d x=\int \frac{\sqrt{33}}{\sqrt{33-33 u^{2}}} d u$

$=\int \frac{1}{\sqrt{1-u^{2}}} d u$

We know that $\int \frac{1}{\sqrt{1-x^{2}}} d x=\sin ^{-1}(x)+c$

$\Rightarrow \int \frac{1}{\sqrt{1-\mathrm{u}^{2}}} \mathrm{du}=\sin ^{-1}(\mathrm{u})$

$=\sin ^{-1}\left(\frac{2 \mathrm{x}-1}{\sqrt{33}}\right)$

Then,

$\Rightarrow \int \frac{x}{\sqrt{-x^{2}+x+8}} d x=\frac{1}{2} \int \frac{2 x-1}{\sqrt{-x^{2}+x+8}} d x-\frac{1}{2} \int \frac{1}{\sqrt{-x^{2}+x+8}} d x$

$=-\sqrt{-x^{2}+x+8}-\frac{1}{2}\left(\sin ^{-1}\left(\frac{2 x-1}{\sqrt{33}}\right)\right)+c$

$\therefore I=\int \frac{x}{\sqrt{-x^{2}+x+8}} d x=-\sqrt{-x^{2}+x+8}-\frac{\sin ^{-1}\left(\frac{2 x-1}{\sqrt{33}}\right)}{2}+c$

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