Question

Solve :
$I=\int \frac{1}{x\left[6\right(\log x{)}^2+3\left(\log x\right)+5]}dx$

Answer 1

Let $logx=t$

$\Rightarrow \frac{1}{x}dx=dt$

$I=\int \frac{dt}{6t^2+3t+5}$

$=\frac{1}{6}\int \frac{dt}{t^2+2\left(\frac{1}{4}\right)\left(t\right)+{\left(\frac{1}{4}\right)}^2+5-{\left(\frac{1}{4}\right)}^2}$

$=\frac{1}{6}\int \frac{dt}{{(t+\frac{1}{4})}^2+{\left(\frac{\sqrt{79}}{4}\right)}^2}$

$=\frac{1}{6}\frac{1}{\left(\frac{\sqrt{79}}{4}\right)}{\tan }^{-1}\left(\frac{t+\frac{1}{4}}{\sqrt{79}/4}\right)+C$

$=\frac{2}{3\sqrt{79}}{\tan }^{-1}\left(\frac{4t+1}{\sqrt{79}}\right)+C$.