MHB Integration Doubt: Answers & Solutions

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The discussion focuses on simplifying a complex integrand involving logarithmic and exponential functions. A proposed simplification reveals a clearer structure, making it easier to analyze. Despite efforts to apply differentiation and substitution methods, further progress in solving the integral remains elusive. Participants express uncertainty about the next steps in the integration process. The conversation highlights the challenges of tackling intricate mathematical expressions effectively.
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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...
 
I also reached this step I tried to create differentiation inside and used substitution too but could not solve it further.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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