Question

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

Answer 1

$I={\int }_{-1}^{1}log\left(\frac{2-x}{2+x}\right)$

$={\int }_{-1}^{1}log(2-x)dx-{\int }_{-1}^{1}log(2+x)dx$ $[\because log\frac{m}{n}=logm-logn]$

$=I_1-I_2$

Let $I_1^{\prime }=\int log(2-x)dx$

Using integration by parts where $u=log(2-x),v=1$

$I_1^{\prime }=log(2-x)\int 1dx-\int \left\{\frac{d}{dx}\right\{log(2-x)\left\}\int 1dx\right\}dx$.

$=xlog(2-x)-\int \frac{-1}{2-x}xdx$

$=xlog(2-x)-\int \frac{2-x-2}{2-x}dx$

$=xlog(2-x)-\int \frac{2-x}{2-x}dx+\int \frac{2}{2-x}dx$

$=xlog(2-x)-x+2\frac{log(2-x)}{-1}$

$=xlog(2-x)-x-2log(2-x)$

$I_1={\int }_{-1}^{1}log(2-x)dx={\left[xlog\right(2-x)-x-2log(2-x\left)\right]}_{-1}^1$

$=1log(2-1)-1-2log(2-1)-(-1)log(2+1)+(-1)+2log(2+1)$

$=log1-1-2log1+log\left(3\right)-1+2log3$

$=-2+3log3$ [since $log1=0$]

$=3log3-2$.

Similarly,

Let $I_2^{\prime }=\int log(2+x)dx$

$=log(2+x)\int 1.dx-\int \left\{\frac{d}{dx}\right\{log(2+x)\left\}\int 1.dx\right\}dx$

$=xlog(2+x)-\int \frac{1}{2+x}xdx$

$=xlog(2+x)-\int \frac{2+x-2}{2+x}dx$

$=xlog(2+x)-\int \frac{2+x}{2+x}dx+\int \frac{2}{2+x}dx$

$=xlog(2+x)-x+2log(2+x)$

$\therefore I_2={\int }_{-1}^{1}log(2+x)dx=\left[xlog\right(2+x)-x+2log(2+x){]}_{-1}^1$

$=1log\left(3\right)-1+2log3-(-1)log1-1-2log1$

$=3log3-2$ $[\because log1=0]$

$\therefore I=I_1-I_2$

$=3log3-2-3log3+2$

$=0$.