Question

If $f\left(x\right)=\underset{1}{\overset{x}{\int }}\frac{\ln t}{1+t}dt,$ then

• A. $f\left(\frac{1}{x}\right)=-\underset{1}{\overset{x}{\int }}\frac{\ln t}{t(1+t)}dt$
• B. $f\left(\frac{1}{x}\right)=\underset{1}{\overset{x}{\int }}\frac{\ln t}{t(1+t)}dt$
• C. $f\left(x\right)+f\left(\frac{1}{x}\right)=0$
• D. $f\left(x\right)+f\left(\frac{1}{x}\right)=\frac{1}{2}(\ln x{)}^2$

Answer 1

Answer: (B), (D)

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

$f\left(x\right)=\underset{1}{\overset{x}{\int }}\frac{\ln t}{1+t}dt$
Replacing $x$ by $\frac{1}{x},$
$f\left(\frac{1}{x}\right)=\underset{1}{\overset{1/x}{\int }}\frac{\ln t}{1+t}dt$
Put $t=\frac{1}{u}\Rightarrow dt=-\frac{1}{u^2}du$
$f\left(\frac{1}{x}\right)=\underset{1}{\overset{x}{\int }}\frac{-\ln u}{\frac{1+u}{u}}\times \frac{-1}{u^2}du=\underset{1}{\overset{x}{\int }}\frac{\ln t}{(1+t)}\cdot \frac{1}{t}dt$

Now, $f\left(x\right)+f\left(\frac{1}{x}\right)=\underset{1}{\overset{x}{\int }}\frac{\ln t}{(1+t)}(1+\frac{1}{t})dt$
$=\underset{1}{\overset{x}{\int }}\frac{\ln t}{t}dt={\left[\frac{(\ln t{)}^2}{2}\right]}_1^x=\frac{(\ln x{)}^2}{2}$