Question

Statement 1 : If $F\left(x\right)={\int }_1^x\frac{logt}{1+t+t^2}dt$, then $F\left(x\right)=-F\left(\frac{1}{x}\right)$
Statement 2 : If $F\left(x\right)={\int }_1^x\frac{logt}{1+t}dt$, then $F\left(x\right)+F\left(\frac{1}{x}\right)=\frac{1}{2}(logx{)}^2$
Which of the following is true:

• A. Statement - 1, Statement -2 are true
• B. Statement - 1, Statement - 2 are false
• C. Statement - 1 is true and Statement - 2 is false
• D. Statement - 1 is false and Statement - 2 is true

Answer 1

The correct option is D Statement - 1 is false and Statement - 2 is true

$F\left(x\right)={\int }_1^{1/x}\frac{logt}{1+t+t^2}dt(putt=\frac{1}{u}\Rightarrow dt=\frac{-1}{u^2}du){\int }_1^x\frac{log\left(\frac{1}{u}\right)}{1+\frac{1}{u}+\frac{1}{u^2}}\left(\frac{-1}{u^2}\right)du={\int }_1^x\frac{logu}{u^2+u+1}du=F\left(x\right)F\left(x\right)+F\left(\frac{1}{x}\right)={\int }_1^x\frac{logt}{t+1}dt+{\int }_1^{1/x}\frac{logt}{t+1}dt(Forsecond,t=\frac{1}{u})={\int }_1^x\frac{logt}{t+1}dt+{\int }_1^x\frac{log\left(\frac{1}{u}\right)}{1+\frac{1}{u}}\left(\frac{-1}{u^2}\right)du={\int }_1^x\frac{logt}{t+1}dt+{\int }_1^x\frac{logt}{t(t+1)}dt={\int }_1^x\frac{logt}{t+1}dt+{\int }_1^x\frac{logt}{t(t+1)}dt={\int }_1^x\frac{logt}{t+1}(1+\frac{1}{t})dt={\int }_1^x\frac{logt}{t}dt=\frac{(logx{)}^2}{2}$