Question

Let $f\left(x\right)={\int }_1^x\frac{\ln t}{1+t}dt$; for $x>0$, then the value of $f\left(e\right)+f\left(\frac{1}{e}\right)$ is

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. none

Answer 1

Answer: (A)

$1$

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

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

So,

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

Using differentiation under integral sign,

$f\left(x\right)=\frac{\ln x}{1+x}+\frac{\ln x}{(1+x)x}$

$=\frac{\ln x.(1+x)}{(1+x).x}$

So, we get

$f\left(x\right)=\frac{(\ln x{)}^2}{2}+C$

Now at $x=1$,

$f\left(1\right)=F\left(1\right)+f\left(\frac{1}{1}\right)=0$

So we get $C=0$

$F\left(x\right)=f\left(x\right)+f\left(\frac{1}{x}\right)=\frac{(\ln x{)}^2}{2}$

Here,

$F\left(e\right)=f\left(e\right)+f\left(\frac{1}{e}\right)=\frac{(\ln e{)}^2}{2}$

$=\frac{1}{2}$