Question

For $x>0$, let $f\left(x\right)={\int }_1^x\frac{\text{log}t}{1+t}dt$. Then $f\left(x\right)+f\left(\frac{1}{x}\right)$ is equal to

• A. $\frac{1}{4}(\log x{)}^4$
• B. $\text{log}x$
• C. $\frac{1}{2}(\log x{)}^2$
• D. $\frac{1}{4}(\log x{)}^2$

Answer 1

The correct option is C

$\frac{1}{2}(\log x{)}^2$

$f\left(\frac{1}{x}\right)={\int }_1^{\frac{1}{x}}\frac{\text{log t}}{1+t}dt$

Put $t=\frac{1}{u}$, so that $dt=(-\frac{1}{u^2})du$, then

$f\left(\frac{1}{x}\right)={\int }_1^x\frac{\text{log}\left(\frac{1}{u}\right)}{1+\frac{1}{u}}(-\frac{1}{u^2})du$

$={\int }_1^x\frac{\text{log u}}{u(1+u)}du={\int }_1^x\frac{\text{log}t}{t(1+t)}dt$

$\therefore f\left(x\right)+f\left(\frac{1}{x}\right)={\int }_1^x(1+\frac{1}{t})\frac{\text{log}t}{1+t}dt$

$={\int }_1^x\frac{\text{log}t}{t}dt$

$=\frac{1}{2}(\text{log}t{)}^2{]}_1^x=\frac{1}{2}(\text{log}x{)}^2$