Prove that the following improper integral $\int_2^\infty x^5 \ln(x^5-1)-5x^5\ln x$ diverges 
Prove that the following improper integral $\int_2^\infty x^5 \ln(x^5-1)-5x^5\ln x$ diverges.

$Proof.$ by simplifying the integral we would get: $$\int _{2}^{\infty }\left( x^{5} \cdot \ln\left( x^{5} -1\right) -5x^{5}\ln( x)\right) dx=\lim _{M\rightarrow \infty }\int _{2}^{M}\left( x^{5}\left(\ln\left( x^{5} -1\right) -5\ln( x)\right)\right) dx\\
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=\lim _{M\rightarrow \infty }\int _{2}^{M}\left( x^{5}\left(\ln\left( x^{5} -1\right) -\ln\left( x^{5}\right)\right)\right) dx=\lim _{M\rightarrow \infty }\int _{2}^{M}\left( x^{5}\ln\left(\frac{x^{5} -1}{x^{5}}\right)\right) dx\\
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=\lim _{M\rightarrow \infty }\int _{2}^{M}\ln\left(\left(\frac{x^{5} -1}{x^{5}}\right)^{x^{5}}\right) dx=\lim _{M\rightarrow \infty }\int _{2}^{M}\ln\left(\left( 1-\frac{1}{x^{5}}\right)^{x^{5}}\right) dx\\
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\xrightarrow[x\rightarrow \infty ]{}\lim _{M\rightarrow \infty }\int _{1}^{M}\ln\left(\left(\frac{1}{e}\right)\right) dx=\lim _{M\rightarrow \infty }\int _{1}^{M} -1\\
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=\lim _{M\rightarrow \infty } -x\Bigl|_{2}^{M} =\lim _{M\rightarrow \infty } -M+2=-\infty $$

Now, I know that it easily diverges because the function has limit $L= \frac{1}{e}$ and it isn't zero. However, is what I have done still valid? or it is a critical mistake?  Thanks!

As you started writing it $$x^5 \log(x^5-1)-5x^5\log( x)=x^5 \log(x^5-1)-x^5\log( x^5)=x^5\log \left(1-\frac{1}{x^5}\right)$$ When $x$ is large
$$x^5\log \left(1-\frac{1}{x^5}\right)=-1-\frac{1}{2 x^5}+O\left(\frac{1}{x^{10}}\right)$$
$$\int x^5\log \left(1-\frac{1}{x^5}\right)\,dx=-x+\frac{1}{8 x^4}+O\left(\frac{1}{x^9}\right)$$
What does happen if you integrate for a large upper bound ?
