(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

[itex]∫^{∞}_{1}[/itex][itex]\frac{dx}{x(2x+5)}[/itex]

2. Relevant equations

Mathematica told me that the answer was [itex]\frac{1}{5}[/itex]nl([itex]\frac{7}{2}[/itex])

3. The attempt at a solution

This is my work.

[itex]∫^{∞}_{1}[/itex][itex]\frac{dx}{x(2x+5)}[/itex] = [itex]lim_{M\rightarrow∞}[/itex][itex]∫^{M}_{1}[/itex][itex]\frac{dx}{x(2x+5)}[/itex] = [itex]lim_{M\rightarrow∞}[/itex][itex]∫^{M}_{1}[/itex][itex]\frac{\frac{1}{5}}{x}[/itex]+[itex]\frac{\frac{-2}{5}}{2x+5}[/itex] = [itex]lim_{M\rightarrow∞}[/itex](([itex]\frac{1}{5}[/itex][itex]∫^{M}_{1}[/itex][itex]\frac{1}{x}[/itex])-([itex]\frac{2}{5}[/itex][itex]∫^{M}_{1}[/itex][itex]\frac{1}{2x+5}[/itex])) = [itex]lim_{M\rightarrow∞}[/itex][itex]\frac{1}{5}[/itex][itex](nl(x)-nl(2x+5))^{M}_{1}[/itex] = [itex]lim_{M\rightarrow∞}[/itex][itex]\frac{1}{5}[/itex]((nl(M)-nl(2M+5))-(nl(1)-nl(7))) = [itex]lim_{M\rightarrow∞}[/itex][itex]\frac{1}{5}[/itex]((nl(M)-nl(2M+5))-(0-nl(7))) = [itex]\frac{1}{5}[/itex]((∞-∞)+nl(7)) = [itex]\frac{1}{5}[/itex]nl(7) .

I'm looking at it over and over again but I just can't seem to figure out where that nl(2) from mathematica came from. Could someone please look at my work and tell me what is wrong with it? I'd greatly appreciate it.

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# Whats wrong with my work for improper integral?

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