Question

For $x>0,$ Let $f\left(x\right)={\int }_1^x\frac{\log t}{1+t}dt$ then find $f\left(x\right)+f\left(\frac{1}{x}\right)$

Answer 1

$\begin{array}{c}f\left(x\right)+f\left(\frac{1}{x}\right)={\int }_1^x\frac{\log tdt}{1+t}+{\int }_1^{\frac{1}{x}}\frac{\log tdt}{1+t}\\ ={\int }_1^x\frac{\log t}{1+t}dt+{\int }_1^x\frac{\log t}{1+t}dt\\ ={\int }_1^x\frac{\log t}{1+t}\left(1+\frac{1}{t}\right)dt\\ ={\int }_1^x\frac{\log t}{1+t}dt\\ =\frac{1}{2}{\left[{\left(\log t\right)}^2\right]}_1^x\\ =\frac{1}{2}{\left(\log x\right)}^2\end{array}$