Mathematics

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


ANSWER

$$0$$


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
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