Mathematics

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

##### SOLUTION
$I=\displaystyle\int^1_{-1}log\left(\dfrac{2-x}{2+x}\right)$
$=\displaystyle\int^1_{-1}log(2-x)dx-\displaystyle\int^1_{-1}log(2+x)dx$ $[\because log\dfrac{m}{n}=log m-log n]$
$=I_1-I_2$
Let $I_1'=\displaystyle\int log (2-x)dx$
Using integration by parts where $u=log (2-x), v=1$
$I_1'=log(2-x)\displaystyle\int 1dx-\displaystyle\int \left\{\dfrac{d}{dx}\{log(2-x)\}\displaystyle\int 1dx\right\}dx$.
$=x log(2-x)-\displaystyle\int \dfrac{-1}{2-x}xdx$
$=x log(2-x)-\displaystyle\int \dfrac{2-x-2}{2-x}dx$
$=x log(2-x)-\displaystyle\int \dfrac{2-x}{2-x}dx+\displaystyle\int \dfrac{2}{2-x}dx$
$=x log(2-x)-x+2\dfrac{log(2-x)}{-1}$
$=x log(2-x)-x-2 log(2-x)$
$I_1=\displaystyle\int^1_{-1}log(2-x)dx=\left[x log(2-x)-x-2 log(2-x)\right]^1_{-1}$
$=1 log(2-1)-1-2 log(2-1)-(-1) log(2+1)+(-1)+2 log(2+1)$
$=log 1-1-2 log 1+log (3)-1+2 log 3$
$=-2+3 log 3$ [since $log 1=0$]
$=3 log 3-2$.
Similarly,
Let $I_2'=\displaystyle\int log (2+x)dx$
$=log (2+x)\displaystyle\int 1.dx-\displaystyle\int \left\{\dfrac{d}{dx}\{log (2+x)\}\displaystyle\int 1.dx\right\}dx$
$=x log(2+x)-\displaystyle\int \dfrac{1}{2+x}xdx$
$=x log (2+x)-\displaystyle\int \dfrac{2+x-2}{2+x}dx$
$=x log(2+x)-\displaystyle\int \dfrac{2+x}{2+x}dx+\displaystyle\int \dfrac{2}{2+x}dx$
$=x log(2+x)-x+2 log(2+x)$
$\therefore I_2=\displaystyle\int^1_{-1}log (2+x)dx=[x log(2+x)-x+2 log(2+x)]^1_{-1}$
$=1 log(3)-1+2log 3-(-1)log 1-1-2 log 1$
$=3 log 3-2$ $[\because log 1=0]$
$\therefore I=I_1-I_2$
$=3 log 3-2-3 log 3+2$
$=0$. Its FREE, you're just one step away

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

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