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

Answer

Let $I =\int \frac{d x}{\sqrt{9+8 x-x^2}}=\int \frac{d x}{\sqrt{9-\left(x^2-8 x\right)}}$
$
\begin{array}{l}
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}}
\end{array}
$
We know that
$
\begin{aligned}
& \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
\end{aligned}
$
So, $I=\sin ^{-1} \frac{x-4}{5}+C$

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