9+x 9-x d x - Vidyadip

Question

$\int \sqrt{\frac{9+x}{9-x}} d x$

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Answer

$\text { Let } I =\int \sqrt{\frac{9+x}{9-x}} d x$
$=\int \sqrt{\frac{9+x}{9-x} \times \frac{9+x}{9+x}} d x$
$=\int \frac{9+x}{\sqrt{(9)^2-x^2}} d x$
$=\int\left[\frac{9}{\sqrt{(9)^2-x^2}}+\frac{x}{\sqrt{(9)^2-x^2}}\right] d x$
$=9 \int \frac{1}{\sqrt{(9)^2-x^2}} d x+\int \frac{x}{\sqrt{(9)^2-x^2}} d x$
$=9 \sin ^{-1}\left(\frac{x}{9}\right)+ I _1$
$\ln I_1 \text {, put }(9)^2- x ^2= t$
$\therefore-2 xdx = dt$
$\therefore xdx =-\frac{1}{2} dt$
$\therefore I _1=-\frac{1}{2} \int \frac{ dt }{\sqrt{ t }}$
$\therefore=-\frac{1}{2} \cdot\left(\frac{ t ^{\frac{1}{2}}}{\frac{1}{2}}\right)+ c$
$=-\sqrt{9^2-x^2}+ c$
$\therefore I =9 \sin ^{-1}\left(\frac{x}{9}\right)-\sqrt{81-x^2}+ c$

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