Question

The function $f\left(x\right)={\int }_0^x\log \left(\frac{1-x}{1+x}\right)dx$

• A. An even function
• B. An odd function
• C. A periodic function
• D. None of these

Answer 1

The correct option is A An even function

$f\left(x\right)={\int }_0^x\log \left(\frac{1-x}{1+x}\right)dx$ ........... $\left(i\right)$

Let $g=\log \left(\frac{1-x}{1+x}\right)$ and $h^{\prime }=1$

Using integration by parts,

$\int g{h}^{\prime }=gh-\int g^{\prime }h$ ....... $\left(ii\right)$

$g^{\prime }=\frac{1+x}{1-x}\frac{d}{dx}\left(\frac{1-x}{1+x}\right)$

$=\frac{1+x}{1-x}\times \left(\frac{-(1+x)-(1-x)}{(1+x{)}^2}\right)$

$=\frac{(1+x)}{(1-x)}\times \left(\frac{-1-x-1+x}{(1+x{)}^2}\right)$

$=\frac{(1+x)}{(1-x)}\times \left(\frac{-2}{(1+x{)}^2}\right)$

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

$\therefore g^{\prime }=\frac{2}{(x+1)(x-1)}$ and $h=x$

Form $\left(ii\right)$, we have

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

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

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

$=x\log \left(\frac{1-x}{1+x}\right)-\log (x+1)-\log (x-1)$

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

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

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

$\therefore f\left(x\right)=x\log \left(\frac{1-x}{1+x}\right)-\log |x^2-1|$

Now, $\therefore f(-x)=-x\log \left(\frac{1+x}{1-x}\right)-\log |x^2-1|$

$=x\log \left(\frac{1-x}{1+x}\right)-\log |x^2-1|=f\left(x\right)$

Thus, $f(-x)=f\left(x\right)$

Hence, $f$ is an even function.