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 $$I_2'=\displaystyle\int log (2+x)dx$$
$$=log (2+x)\displaystyle\int 1.dx-\displaystyle\int \left\{\dfrac{d}{dx}\{log (2+x)\}\displaystyle\int 1.dx\right\}dx$$
$$=x log(2+x)-\displaystyle\int \dfrac{1}{2+x}xdx$$
$$=x log (2+x)-\displaystyle\int \dfrac{2+x-2}{2+x}dx$$
$$=x log(2+x)-\displaystyle\int \dfrac{2+x}{2+x}dx+\displaystyle\int \dfrac{2}{2+x}dx$$
$$=x log(2+x)-x+2 log(2+x)$$
$$\therefore I_2=\displaystyle\int^1_{-1}log (2+x)dx=[x log(2+x)-x+2 log(2+x)]^1_{-1}$$
$$=1 log(3)-1+2log 3-(-1)log 1-1-2 log 1$$
$$=3 log 3-2$$ $$[\because log 1=0]$$
$$\therefore I=I_1-I_2$$
$$=3 log 3-2-3 log 3+2$$
$$=0$$.
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Subjective Medium Published on 17th 09, 2020
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