Explanation
Consider the given integral.
$$ I=\int_{1}^{e}{\dfrac{dx}{\ln \left( {{x}^{x}}\cdot {{e}^{x}} \right)}} $$
$$ I=\int_{1}^{e}{\dfrac{dx}{\ln {{x}^{x}}+\ln {{e}^{x}}}} $$
$$ I=\int_{1}^{e}{\dfrac{dx}{x\ln x+x\ln e}} $$
$$ I=\int_{1}^{e}{\dfrac{dx}{x\ln x+x}} $$
$$ I=\int_{1}^{e}{\dfrac{dx}{x\left( \ln x+1 \right)}} $$
Let $$t=\ln x+1$$
$$ \dfrac{dt}{dx}=\dfrac{1}{x}+0 $$
$$ dt=\dfrac{dx}{x} $$
Therefore,
$$ I=\int_{1}^{2}{\dfrac{dt}{t}} $$
$$ I=\left[ \ln \left( t \right) \right]_{1}^{2} $$
$$ I=\ln \left( 2 \right)-\ln \left( 1 \right) $$
$$ I=\ln 2 $$
Hence, this is the answer.
Please disable the adBlock and continue. Thank you.