Question

$f\left(x\right)={\int }_0^{x}ln\left(\frac{1-t}{1+t}\right)dt\Rightarrow$

• A. $f\left(x\right)$ is an even function
• B. $f\left(x\right)$ is an odd function
• C. $f\left(x\right)$ is neither even nor odd
• D. $f\left(x\right)$ cannot be a function

Answer 1

The correct option is A $f\left(x\right)$ is an even function

$f\left(x\right)={\int }_0^x\ln \left(\frac{1-t}{1+t}\right)dt$

$f(-x)={\int }_0^{-x}\ln \left(\frac{1-t}{1+t}\right)dt$

putting $t=-y$

$dt=-dy$

$f(-x)=-{\int }_0^x\ln \left(\frac{1+y}{1-y}\right)dy$

$f(-x)={\int }_0^x\ln \left(\frac{1-y}{1+y}\right)dy$

$f(-x)={\int }_0^x\ln \left(\frac{1-t}{1+t}\right)dt$

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

So, $f\left(x\right)$ is an even function.