Prove
Question:

$\frac{1}{\sqrt{8+3 x-x^{2}}}$

Solution:

$8+3 x-x^{2}$ can be written as $8-\left(x^{2}-3 x+\frac{9}{4}-\frac{9}{4}\right)$

Therefore,

$8-\left(x^{2}-3 x+\frac{9}{4}-\frac{9}{4}\right)$

$=\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}$

$\Rightarrow \int \frac{1}{\sqrt{8+3 x-x^{2}}} d x=\int \frac{1}{\sqrt{\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}}} d x$

Let $x-\frac{3}{2}=t$

$\therefore d x=d t$

$\Rightarrow \int \frac{1}{\sqrt{\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}}} d x=\int \frac{1}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^{2}-t^{2}}} d t$

$=\sin ^{-1}\left(\frac{t}{\frac{\sqrt{41}}{2}}\right)+\mathrm{C}$

$=\sin ^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\right)+\mathrm{C}$

$=\sin ^{-1}\left(\frac{2 x-3}{\sqrt{41}}\right)+\mathrm{C}$

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