Question

Let F(x) = f(x) + $f\left(\frac{1}{x}\right)$, where f(x) = ${\int }_1^x\frac{logt}{1+t}dt$ . Then F(e) =

• A. $\frac{1}{2}$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is A $\frac{1}{2}$

$Givenf\left(x\right)={\int }_1^x\frac{logt}{1+t}dt.Putt=\frac{1}{z}\Rightarrow dt=-\frac{1}{z^z}dz$
$f\left(\frac{1}{x}\right)={\int }_1^{\frac{1}{x}}\frac{logt}{1+t}dt={\int }_1^x\frac{log\left(\frac{1}{z}\right)}{1+\left(\frac{1}{z}\right)}\left(\frac{-1}{z^2}\right)dz={\int }_1^x\frac{logz}{z(z+1)}dz={\int }_1^x\frac{logt}{t(t+1)}dt$
$F\left(x\right)=f\left(x\right)+f\left(\frac{1}{x}\right)={\int }_1^x[\frac{logt}{1+t}+\frac{logt}{t(t+1)}]dt$
$={\int }_1^x\frac{logt}{t}dt={\left[\frac{(logt{)}^2}{t}\right]}_1^x=\frac{(logx{)}^2}{2}\Rightarrow F\left(e\right)=\frac{1}{2}$