Question

Evaluate
$\int \frac{1}{x}\{\log e^{ex}.\log e^{e^{2}x}.\log e^{e^{3}x}\}dx$

Answer 1

$I=\int \frac{1}{x}\{\log e^{ex}.\log e^{e^{2}x}.\log {}_e{e}^{3}x\}dx$
$=\int \frac{1}{x}\{\frac{1}{\log e^{ex}}.\frac{1}{\log e^{e^{2}x}}.\frac{1}{\log {}_e{e}^{3}x}\}dx$
$=\int \frac{dx}{x\{\log e^{ex}+\log e^x\}\{\log e^{e^2}+\log e^{x^2}\}\{\log e^{e^2}+\log x^2\}}$
$=\int \frac{dx}{x\{1+\log e^x\}\{2+\log e^x\}\{3+\log e^2\}}$
Put $\log e^x=t\Rightarrow \frac{1}{x}=dt$
$I=\int \frac{dt}{(1+t)(2+t)(3+t)}=\int (\frac{1}{2}.\frac{1}{(1+t)}-\frac{1}{(2+t)}+\frac{1}{(3+t)})dt$
$=\frac{1}{2}\log |1+t|-\log |2+t|+\log |3+t|+C$.
$=\frac{1}{2}\log |1+\log e^x|-\log |2+\log e^x|+\log |3+\log e^x|+C$.