Question

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

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

Answer 1

The correct option is D $\frac{1}{2}(\log x{)}^2$

$f\left(x\right)=\underset{1}{\overset{x}{\int }}\frac{\log t}{1+t}dtf\left(\frac{1}{x}\right)=\underset{1}{\overset{\frac{1}{x}}{\int }}\frac{\log t}{1+t}dt\text{Put}t=\frac{1}{z}\Rightarrow dt=-\frac{1}{z^2}dzf\left(\frac{1}{x}\right)=\underset{1}{\overset{x}{\int }}\frac{\log \frac{1}{z}}{1+\frac{1}{z}}(-\frac{1}{z^2})dz=\underset{1}{\overset{x}{\int }}\frac{\log z}{z(1+z)}dz=\underset{1}{\overset{x}{\int }}\frac{\log t}{t(1+t)}dt\Rightarrow f\left(x\right)+f\left(\frac{1}{x}\right)=\underset{1}{\overset{x}{\int }}\frac{\log t}{t}dt=\frac{(\log x{)}^2}{2}$