Mathematics

# Solve $\int _{ 0 }^{ 1 }{ \log\left( \dfrac { 2-x }{ 2+x} \right) } dx$

##### SOLUTION
$I=\displaystyle\int^1_)log \left(\dfrac{2-x}{2+x}\right)dx$
$=\displaystyle\int^1_0log (2-x)dx-\displaystyle\int^1_0log (2+x)dx$ $[\because log \dfrac{m}{n}= log m- log n]$
$=I_1-I_2$
Let $I_1'=\displaystyle\int log(2-x)dx$ [using integration by parts]
$=log (2-x)\displaystyle\int 1.dx-\displaystyle\int \left\{\dfrac{d}{dx}(log (2-x))\displaystyle\int 1.dx\right\}dx$
$=x log(2-x)-\displaystyle\int \dfrac{1}{2-x}.(-1)xdx$
$=x log(2-x)+\displaystyle\int \dfrac{x}{2-x}dx$
$=x log(2-x)+\displaystyle\int \dfrac{-(2-x)+2}{2-x}dx$
$=x log(2-x)+\displaystyle\int \dfrac{-(2-x)}{2-x}dx+\displaystyle\int \dfrac{2}{2-x}dx$
$=x log(2-x)-\displaystyle\int dx+2\displaystyle\int \dfrac{dx}{2-x}$
$=x log(2-x)-x-2 log(2-x)$
$\therefore I_1=\displaystyle\int^1_0log(2-x)dx=[x log(2-x)-x-2 log(2-x)]^1_0$
$=1.log(2-1)-1-2 log(2-1)-0+0+2 log 2$
$=log 1-1-2 log 1+2 log 2$
$=-1+2 log 2$ $[\because log 1=0]$
$=2 log 2-1$
Similarly $I_2'=\displaystyle\int log (2+x)dx$
$=log (2+x)\displaystyle\int dx-\displaystyle\int \left\{\dfrac{d}{dx}(log (2+x))\displaystyle\int dx\right\}dx$
$=x log(2+x)-\displaystyle\int \dfrac{1}{2+x}xdx$
$=x log(2+x)-\displaystyle\int \dfrac{2+x-2}{2+x}dx$
$=x log(2+x)-\displaystyle\int \dfrac{2+x}{2+x}dx+\displaystyle\int \dfrac{2}{2+x}dx$
$=x log(2+x)-\displaystyle\int dx+2\displaystyle\int \dfrac{1}{2+x}dx$
$=x log(2+x)-x+2 log(2+x)$
$I_2=\displaystyle\int^1_0log(2+x)dx$
$=[x log(2+x)-x+2log (2+x)]^1_0$
$=1 log(2+1)-1+2 log(2+1)-0+0-2 log(2)$
$=log(3)-1+2 log 3-2 log 2$
$=3 log 3-2 log 2-1$
$\therefore I=I_1-I_2$
$=2log 2-1-3log 3+2log 2+1$
$=4 log 2-3 log 3$
$=log 2^4-log 3^3=log \dfrac{2^4}{3^3}=log\left(\dfrac{16}{27}\right)$.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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