Mathematics | TS EAMCET 2019 | Discussion Page

$\int\frac{dx}{(1+x)\sqrt{8+7x-x^{2}}}$ =

$-\frac{2}{9}\sqrt{\frac{8-x}{1+x}}+c$

$-\frac{1}{9}\sqrt{\frac{1+x}{8-x}}+c$

$-\frac{2}{9}\sqrt{\frac{1+x}{8-x}}+c$

$\frac{2}{9}\sqrt{\frac{8+x}{1+x}}+c$

Option A

We have

I= $\int\frac{dx}{(1+x)\sqrt{8+7x-x^{2}}}$

$I=\int\frac{dx}{(1+x)\sqrt{(8-x)(1+x)}}$

$I=\int\frac{dx}{\sqrt{\frac{8-x}{1+x}(1+x)^{2}}}$

Put $\frac{8-x}{1+x}=t^{2}\Rightarrow\left(\frac{(1+x)(-1)-(8-x)}{(1+x)^{2}}\right)dx=2t dt$

$\Rightarrow$ $\frac{-9}{(1+x)^{2}}dx=2t dt$

$\therefore$ $I=\frac{-2}{9}\int\frac{tdt}{t}$

$I =\frac{-2}{9}\int dt=-\frac{2}{9}t+c$

$\therefore$ I= $-\frac{2}{9}\sqrt{\frac{8-x}{1+x}}+c$

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