Question

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

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

Answer 1

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

$f\left(x\right)={\int }_1^x\frac{logt}{1+t}dt.--------\left(1\right)$
$f\left(\frac{1}{x}\right)={\int }_1^{\frac{1}{x}}\frac{logt}{1+t}dt$
put $t=\frac{1}{y}$
$\Rightarrow dt=-\frac{1}{y^2}dy---------\left(2\right)$
$Add\left(1\right)and\left(2\right)$
$F\left(e\right)=f\left(x\right)+f\left(\frac{1}{x}\right)={\int }_1^e\frac{logt}{1+t}dt+{\int }_1^e\frac{logt}{(1+t)t}dt$
$={\int }_1^e\frac{logt}{t}dt={\left[\frac{(logt{)}^2}{2}\right]}_1^e=\frac{1}{2}$