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$\displaystyle \int \frac{\ln x}{x^2} \mathrm dx$

I just can't seem to figure this one out. I tried integrating by parts but I'm stuck.

Git Gud
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Paze
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    Letting $u=\ln x$ and $dv=1/x^2 dx$ seems to make the problem rather tractable. – Hayden Nov 24 '13 at 22:25
  • Yes but the derivative of $lnx$ is not $\frac{1}{x^2}$ so I don't understand how I'd use substitution here. I need another variable to make it work it seems. – Paze Nov 24 '13 at 22:26
  • It seems like you're confusing integration by parts for u-substitution. – Hayden Nov 24 '13 at 22:29
  • Ah! Why the heck didn't I get this problem. I thought I had tried both ways with both functions as u.

    Thanks.

     – Paze Nov 24 '13 at 22:39

1 Answers1

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Integration by parts is not u-substitution. You split the integrand into two parts, $u$ and $dv$. Choose $u$ so that it will simplify when you take it's derivative. In this case, $u$ simplifies when it is ln(x) and the derivative is 1/x .

$u = \ln(x)$

$du = (1/x)dx$

$v = -1/x$

$dv = (1/(x^2))dx$

$uv - \int{vdu} = (\ln(x))\cdot(-1/x)- \int{(-1/(x^2)) dx} = (-\ln x)/x - (1/x) = -( ln x+1)/x$

Marcus Johnson
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  • Just putting formulas might answer the question, but I don't think that it helps to address the confusion. Perhaps talk about why you choose these functions for your $u$ and $v$ as well as why this isn't $u$-substitution might help the person who asked the question more. – Hayden Nov 24 '13 at 22:34
  • The answer's correct and it can be taken as hints for the OP to find out what's going on, as (s)he already knows what is integration by parts. The downvote is just unfair and out of place, imo. +1 – DonAntonio Nov 24 '13 at 22:39
  • Thank you. I understand my error now. – Paze Nov 24 '13 at 22:41