Question

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

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

Answer 1

Answer: (C)

$\frac{1}{2}(logx{)}^2$

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

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

Put t=1/u

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

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

Add (1) and (2)

$={\int }_1^x\frac{\log u}{u}du$

$z=logudz=\frac{du}{u}$

${\int }_1^{x}zdz$

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