DaalChawal
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How to integrate x^2 times e^(ax) using integration by parts
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Prove It
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MHB
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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...
$\displaystyle \begin{align*} \frac{\ln{\left( \mathrm{e}\,x^{x+1} \right)} + \left[ \ln{ \left( x^{\sqrt{x}} \right) } \right]^2 }{1 + x\ln{ \left( x \right) } \ln{ \left( \mathrm{e}^x\,x^x \right) }} &= \frac{ \ln{\left( \mathrm{e} \right) } + \ln{ \left( x^{x+1} \right) } + \left[ \sqrt{x} \, \ln{ \left( x \right) } \right] ^2 }{ 1 + x\ln{ \left( x \right) } \left[ \ln{\left( \mathrm{e}^x \right) } + \ln{ \left( x^x \right) } \right] } \\ &= \frac{ 1 + \left( x + 1 \right) \ln{ \left( x \right) } + x \, \left[ \ln{ \left( x \right) } \right] ^2 }{ 1 + x \ln{ \left( x \right) } \left[ x + x\ln{ \left( x \right) } \right] } \end{align*}$
I don't know if this helps, but it looks simpler at least...
DaalChawal
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I also reached this step I tried to create differentiation inside and used substitution too but could not solve it further.