Question

The integral ${\int }_{-1/2}^{1/2}\left(\right[x]+1n\left(\frac{1+x}{1-x}\right))dx$ equal to:

Answer 1

$\int (x{f}^{{}^{\prime }}\left(x\right)+f\left(x\right))dx=xf\left(x\right)+c\Rightarrow \int (x\left(\frac{1}{x}\right)+\ln x)dx=x\ln x+c$

i.e, $\int \ln xdx=x\ln x-\int dx=x\ln x-x$

$\therefore {\int }_{-1/2}^{1/2}\ln \frac{1+x}{1-x}dx={\int }_{-1/2}^{1/2}(\ln (1+x)-\ln (1-x))dx$

$=|(x+1)\ln (x+1)+(1-x)\ln (1-x)-(1+x+1-x){|}_{-1/2}^{1/2}=0$

and ${\int }_{-1/2}^{1/2}\left[x\right]dx={\int }_{-1/2}^0(-1)dx+{\int }_0^{1/2}0dx=(-x){\}}_{-1/2}^0=\frac{(-1)}{2}$

Then ${\int }_{-1/2}^{1/2}(\left[x\right]+\ln \frac{1+x}{1-x})dx=\frac{(-1)}{2}+0=\frac{(-1)}{2}$