Question

If $f\left(x\right)={\int }_1^x\frac{\log t}{t+1}dt$ and $f\left(x\right)+f\left(1/x\right)=k{\left(\log x\right)}^2$, then $k$ equal to

• A. $1$
• B. $1/2$
• C. $1/4$
• D. $1/3$

Answer 1

The correct option is B $1/2$

$f\left(x\right)={\int }_1^x\frac{\log t}{t+1}dt$
$f\left(\frac{1}{x}\right)={\int }_0^{\frac{1}{x}}\frac{\log t}{1+t}dt={\int }_1^x\left(\log u\right)(\frac{1}{u}-\frac{1}{u+1})du$
$={\int }_1^x\frac{\log u}{u}du-{\int }_1^x\frac{\log u}{u+1}du={\int }_1^x\frac{\log t}{t}dt-{\int }_1^x\frac{\log t}{t+1}dt$
Now $f\left(x\right)+f\left(1/x\right)=k{\left(\log x\right)}^2$
$\Rightarrow {\int }_1^x\frac{\log t}{t}dt=k{\left(\log x\right)}^2\Rightarrow \frac{1}{2}{\left(\log x\right)}^2=k{\left(\log x\right)}^2$