MHB Integration Doubt: Answers & Solutions

[DaalChawal]

 

[Prove It]

It might be worth simplifying the integrand...

$\displaystyle \begin{align*} \frac{\ln{\left( \mathrm{e}\,x^{x+1} \right)} + \left[ \ln{ \left( x^{\sqrt{x}} \right) } \right]^2 }{1 + x\ln{ \left( x \right) } \ln{ \left( \mathrm{e}^x\,x^x \right) }} &= \frac{ \ln{\left( \mathrm{e} \right) } + \ln{ \left( x^{x+1} \right) } + \left[ \sqrt{x} \, \ln{ \left( x \right) } \right] ^2 }{ 1 + x\ln{ \left( x \right) } \left[ \ln{\left( \mathrm{e}^x \right) } + \ln{ \left( x^x \right) } \right] } \\ &= \frac{ 1 + \left( x + 1 \right) \ln{ \left( x \right) } + x \, \left[ \ln{ \left( x \right) } \right] ^2 }{ 1 + x \ln{ \left( x \right) } \left[ x + x\ln{ \left( x \right) } \right] } \end{align*}$

I don't know if this helps, but it looks simpler at least...

 

[DaalChawal]

I also reached this step I tried to create differentiation inside and used substitution too but could not solve it further.