Mathematics

# $\overset { 4 }{ \underset { -4 }{ \displaystyle\int } }\log \left( \dfrac { 9-x }{ 9+x } \right)$ dx equals.

$0$

##### SOLUTION
$I=\displaystyle\int^4_{-4}log\left(\dfrac{9-x}{9+x}\right)dx$ …………$(1)$
$I(x)\displaystyle\int^b_a=I_{(a+b-x)}\displaystyle\int^b_a$
$I=\displaystyle\int^4_{-4}log\left(\dfrac{9-(4-4-x)}{9-(4-4-x)}\right)$
$=\displaystyle\int^4_{-4}log\left(\dfrac{9+x}{9-x}\right)dx$ …………$(2)$
$(1)+(2)$
$\Rightarrow 2I=\displaystyle\int^4_{-4}\left[log\left(\dfrac{9-x}{9+x}\right)+log\left(\dfrac{9+x}{9-x}\right)\right] dx$
$=\displaystyle\int^4_{-4}log\left(\dfrac{9-x}{9+x}\right)\left(\dfrac{9+x}{9-x}\right)dx$
$=\displaystyle\int^4_{-4}log 1=0$
$\Rightarrow I=0$
$\therefore \displaystyle\int^4_{-4}log\left(\dfrac{9-x}{9+x}\right)dx=0$.

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
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