Question
Evaluate $\int \frac{d x}{\sqrt{9+8 x-x^2}}$.

Answer

Let $I =\int \frac{d x}{\sqrt{9+8 x-x^2}}$
$=\int \frac{d x}{\sqrt{9-\left(x^2-8 x\right)}}$
$I=\int \frac{d x}{\sqrt{9-\left(x^2-8 x+16\right)+16}}$
$=\int \frac{d x}{\sqrt{25-(x-4)^2}}$
$I=\int \frac{d x}{\sqrt{(5)^2-(x-4)^2}}$
We know that
$\int \frac{1}{\sqrt{a^2-x^2}} d x$
$=\sin ^{-1} \frac{x}{a}+C$
$\therefore I=\int \frac{d x}{\sqrt{(5)^2-(x-4)^2}}$
$=\sin ^{-1} \frac{x-4}{5}+C$
So $, I=\sin ^{-1} \frac{x-4}{5}+C$

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