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I was trying to find the following indefinite integral: $$\int \frac{\ln(x)^2}{x}\,dx$$ Using u-substitution, I was getting two different answers based on if I took the exponent on the natural log out of the integral:

Taking it out $$2 \int \frac{\ln(x)}{x}\,dx$$ $$u=\ln(x)$$ $$du=\frac{1}{x}\,dx$$ $$\Rightarrow 2\int u \, du=2\,\Big(\frac{u^2}{2}\Big)+c$$ $$\boldsymbol{=2\ln(x)+c}$$

Not taken out $$\int \frac{\ln(x)^2}{x}\,dx=\int u^2\,du=\frac{1}{3}\,u^3+c$$ $$=\boldsymbol{\ln(x)+c}$$

Cannot quite figure out why I'm getting two different answers. Maybe it has to do with something I haven't yet learnt in class. Any insight would be appreciated.

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    $\begingroup$ please note $(\ln(x))^2$ is not the same as $\ln(x^2)=2\ln (x)$. I haven't read carefully but this could certainly be the source of your error $\endgroup$
    – operatorerror
    Dec 10, 2017 at 3:01
  • $\begingroup$ Ah I see. That was my mistake. Thank you for you answer! $\endgroup$
    – Hamza Qayyum
    Dec 10, 2017 at 3:02
  • $\begingroup$ The solution is $\dfrac{ln(x)^3}{3}$. $\endgroup$
    – QFi
    Dec 10, 2017 at 3:06
  • $\begingroup$ the integrals you are computing are not definite $\endgroup$
    – Masacroso
    Dec 10, 2017 at 3:15

2 Answers 2

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Both are wrong. The first fails because $\ln(x)^2 \neq 2 \ln(x)$. Then when you backsubstituted you lost the square on $u$. The second fails because you lost the square of the log. You put it back in in the $\int u^2$, but then when you backsubstituted you lost the cube. $u^3=(\ln (x))^3$ The correct answer is $\frac 13(\ln(x))^3+c$

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  • $\begingroup$ Yep, that was it. Thank you for your answer! Filled a hole in my math knowledge. $\endgroup$
    – Hamza Qayyum
    Dec 10, 2017 at 3:09
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$$ \int \Big(\! \ln x\Big)^2 \left( \frac {dx} x \right) = \int u^2 \, du. $$ $$\text{Note that } \ln(x^2) = 2\ln x \text{ but } \left( \ln x\right)^2 \ne 2\ln x.$$

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