Question

If $f(x)={\int }_1^x\frac{\log t}{1+t+t^2}dx,\forall x\ge 1$ then $f(x)$

• A. $f\left(\frac{1}{x}\right)$
• B. $f\left(\frac{1}{x^2}\right)$
• C. $f\left(x^2\right)$
• D. $-f\left(\frac{1}{x}\right)$

Answer 1

Answer: (A)

Given $f(x)=\underset{1}{\overset{x}{\int }}\frac{\log t}{1+t+t^2}dx$
$\Rightarrow f\left(\frac{1}{x}\right)=\underset{1}{\overset{1/x}{\int }}\frac{\log t}{1+t+t^2}dx$
Let $x=\frac{1}{t}$
$\Rightarrow dx=\frac{-dt}{t^2}$
$\Rightarrow f\left(\frac{1}{x}\right)=\underset{1}{\overset{x}{\int }}\frac{\log \frac{1}{t}}{1+\frac{1}{t}+\frac{1}{t^2}}\left(-\frac{1}{t^2}\right)dt$
$=\underset{1}{\overset{x}{\int }}\frac{\log t}{1+t+t^2}dt=f(x)$