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Q.
If $f(x)=\int_{1}^{x} \frac{\log t}{1+t+t^{2}} d x, \forall x \geq 1$ then $f(x)$
Integrals
Solution:
Given $f(x)=\int\limits_{1}^{x} \frac{\log t}{1+t+t^{2}} d x$
$\Rightarrow f\left(\frac{1}{x}\right)=\int\limits_{1}^{1 / x} \frac{\log t}{1+t+t^{2}} d x$
Let $x=\frac{1}{t}$
$\Rightarrow d x=\frac{-d t}{t^{2}}$
$\Rightarrow f\left(\frac{1}{x}\right)=\int\limits_{1}^{x} \frac{\log \frac{1}{t}}{1+\frac{1}{t}+\frac{1}{t^{2}}}\left(-\frac{1}{t^{2}}\right) d t$
$=\int\limits_{1}^{x} \frac{\log t}{1+t+t^{2}} d t=f(x)$