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I'm trying to integrate $\int r(\ln r)^2\, dr$.

I let $u=\ln r$ and $du=\dfrac 1 r \, dr$. So then I get, $\displaystyle\int \dfrac{ru^2} r \, du = \int u^2 \, du = \dfrac 1 3 (u)^3+C$$$=\frac 1 3 (\ln r)^3 +C$$

Now wolfram alpha does the problem via integration by parts and gives the answer, $$\dfrac{1}{4}r^2(2\log^2(r)-2\log(r)+1)+C$$

I'm struggling to understand why I can't use substitution to solve the integral.

Thanks.

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    your result is right, you also differentiate your result – 2017-01-18
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    @Dr.Sonnhard Graubner how is Op's result right – 2017-01-18
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    @Dr.SonnhardGraubner You sure it's right? Differentiating it does not give the original function... – 2017-01-18
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    or you have made a mistake or i or we both – 2017-01-18

4 Answers 4

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Because you do it wrong: By $du = \dfrac 1 r\,dr$ we get $\int r^2 u^2 \, du$ which does not help at first sight.

So a little trick is needed. We get $r\,du = dr$ and so we can use $u = \log r \Leftrightarrow e^u = r$ which yields the integral $$\int u^2e^{2u} \, du$$ which easily can be solved by integration by parts.

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    Actually $\displaystyle \int r^2 u^2 \, du$ does help: just realize that $r^2=(e^u)^2 = e^{2u}. \qquad$ – 2017-01-18
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    @MichaelHardy Yes, but I think this is not entirely obvious, especially when you've never seen this before. But for that I've added the second part of my answer. – 2017-01-18
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You've done the substitution wrong. To make it more obvious, let $r=e^u$ so that $dr=e^u\ du$.

$$=\int e^uu^2(e^u\ du)=\int u^2e^{2u}\ du$$

One can then solve the following:

$$f(x)=\int e^{xu}\ du$$

So that

$$f''(2)=\int u^2e^{2u}\ du$$

which avoids integration by parts :)

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It gets converted to to $u^2.e^{2u} $ you have made a mistake there rather make . Now apply by parts twice with $u^2=x $ and $e^{2u}=v $ to get the result.

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I would recommend attempting to redo your substitution. Since in your substitution you exchanged $dr$ for $du/r$, while in your calculation you actually showed $dr = r \, du$