Question

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

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

Answer 1

Answer: (B)

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

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

$Now$

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

$Lett=\frac{1}{y}\Rightarrow dt=\frac{-1}{y^2}dy$

$f\left(\frac{1}{x}\right)={\int }_1^x\frac{\log \frac{1}{y}}{1+\frac{1}{y}}\times -\frac{1}{y^2}dy$ (replacing $t$ with $\frac{1}{y}$ )

$as,t=\frac{1}{y}\Rightarrow \frac{1}{x}=\frac{1}{y}\Rightarrow x=y$ (upper limit)

$={\int }_1^x\frac{-\log y}{1+y}\times y\times -\frac{1}{y^2}dy$

$={\int }_1^x\frac{\log y}{1+y}\times \frac{1}{y}dy$

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

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

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

$putting\log t=z$

$\frac{1}{t}dt=dz$

${\int }_0^{\log x}zdz$...... $\left(upperlimitbecomeslogxsimilarly\right)$

$={{\left[\frac{z^2}{2}\right]}_0}^{\log x}=\frac{1}{2}{\left(\log x\right)}^2$