Question

Solve ${\int }_0^1\log \left(\frac{2-x}{2+x}\right)dx$

Answer 1

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

$={\int }_0^{1}log(2-x)dx-{\int }_0^{1}log(2+x)dx$ $[\because log\frac{m}{n}=logm-logn]$

$=I_1-I_2$

Let $I_1^{\prime }=\int log(2-x)dx$ [using integration by parts]

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

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

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

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

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

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

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

$\therefore I_1={\int }_0^{1}log(2-x)dx=\left[xlog\right(2-x)-x-2log(2-x){]}_0^1$

$=1.log(2-1)-1-2log(2-1)-0+0+2log2$

$=log1-1-2log1+2log2$

$=-1+2log2$ $[\because log1=0]$

$=2log2-1$

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

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

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

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

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

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

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

$I_2={\int }_0^{1}log(2+x)dx$

$=\left[xlog\right(2+x)-x+2log(2+x){]}_0^1$

$=1log(2+1)-1+2log(2+1)-0+0-2log\left(2\right)$

$=log\left(3\right)-1+2log3-2log2$

$=3log3-2log2-1$

$\therefore I=I_1-I_2$

$=2log2-1-3log3+2log2+1$

$=4log2-3log3$

$=log{2}^4-log{3}^3=log\frac{2^4}{3^3}=log\left(\frac{16}{27}\right)$.