$$\int\frac{\ln x-2}{x\sqrt{\ln x}} \, dx$$
Could I ask you, please, for helping me out with this example? I have some issues to get an result of this… Thank you in advance
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Have you tried doing this by simple substitution? – Kevin May 04 '14 at 20:43
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Yes, I am actually asked to do it with substitution, but I don't know what's just wrong... – user2932090 May 04 '14 at 20:45
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Make sure you also substitute $dx$ with $du$ if you substitute $x$ with $u$. – Kevin May 04 '14 at 20:46
4 Answers
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$$ \int\frac{\ln x - 2}{\sqrt{\ln x}} \underbrace{{}\ \left( \frac{dx}{x} \right)\ {}}_\text{HINT} $$
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We have
$$\int\frac{\ln x-2}{x\sqrt{\ln x}}dx=\int\frac{\sqrt{\ln x}}{x}dx-\int\frac{2}{x\sqrt{\ln x}}dx=\frac23(\ln x)^{3/2}-4\sqrt{\ln x}+C$$
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I checked the correct results and there's this: $$2lnx(\frac{\ln x}{3} - 2) + C$$ -- however, I am still not able to get there. – user2932090 May 12 '14 at 18:29
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Let $y = \sqrt{\ln x} \to \ln x = y^2 \to x = e^{y^2} \to dx = 2y\cdot e^{y^2}\,dy$. So your integral equals to: $$\int 2y^2 - 4 \,dy = \dfrac{2y^3}{3} - 4y + C = \dfrac{2(\ln x)^{\frac{3}{2}}}{3} - 4\sqrt{\ln x} + C$$
DeepSea
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I changed several occurrences of $lnx$ to $\ln x$. That is standard usage. ${}\qquad{}$ – Michael Hardy May 04 '14 at 20:53
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$t= \sqrt{\ln x}$, then $dt = (\frac{d}{dx}\sqrt{\ln x}) dx = \frac{1}{2x\sqrt{\ln x}} dx $
So your integral becomes \begin{equation*} 2\int (t^2 - 2) dt \end{equation*}
AnalysisStudent0414
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