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$\int \frac{d x}{(1+x) \sqrt{8+7 x-x^2}}=$
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2198 Upvotes
Verified Answer
The correct answer is:
$-\frac{2}{9} \sqrt{\frac{8-x}{1+x}}+c$
We have,
$$
\begin{aligned}
& I=\int \frac{d x}{(1+x) \sqrt{8+7 x-x^2}} \\
& I=\int \frac{d x}{(1+x) \sqrt{(8-x)(1+x)}} \\
& I=\int \frac{d x}{\sqrt{\frac{8-x}{1+x}}(1+x)^2} \\
& \operatorname{Put} \frac{8-x}{1+x}=t^2 \Rightarrow\left(\frac{(1+x)(-1)-(8-x)}{(1+x)^2}\right) d x=2 t d t \\
& \Rightarrow \quad \frac{-9}{(1+x)^2} d x=2 t d t
\end{aligned}
$$
$$
\begin{array}{rlrl}
\therefore & I & =\frac{-2}{9} \int \frac{t d t}{t} \\
& I=-\frac{2}{9} \int d t=-\frac{2}{9} t+c \\
\therefore & I & =\frac{-2}{9} \sqrt{\frac{8-x}{1+x}}+c
\end{array}
$$
$$
\begin{aligned}
& I=\int \frac{d x}{(1+x) \sqrt{8+7 x-x^2}} \\
& I=\int \frac{d x}{(1+x) \sqrt{(8-x)(1+x)}} \\
& I=\int \frac{d x}{\sqrt{\frac{8-x}{1+x}}(1+x)^2} \\
& \operatorname{Put} \frac{8-x}{1+x}=t^2 \Rightarrow\left(\frac{(1+x)(-1)-(8-x)}{(1+x)^2}\right) d x=2 t d t \\
& \Rightarrow \quad \frac{-9}{(1+x)^2} d x=2 t d t
\end{aligned}
$$
$$
\begin{array}{rlrl}
\therefore & I & =\frac{-2}{9} \int \frac{t d t}{t} \\
& I=-\frac{2}{9} \int d t=-\frac{2}{9} t+c \\
\therefore & I & =\frac{-2}{9} \sqrt{\frac{8-x}{1+x}}+c
\end{array}
$$
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