How to evaluate the following integral?

$$\int \frac {\ln(3x+7)}{x^2}\,\mathrm dx$$

I tried substituting both $u = 3x+7$ and $u = \ln(3x+7)$, but the resulting integral seems to be much more complex. I also considered integration by parts, but didn't start because it seemed useless in this case.

The substitution $t = -\dfrac 1x$ was promising, because I could get rid of $\dfrac 1{x^2}$ and remained with $\ln\left(-\dfrac3t+7\right)$, but then I did not know how to proceed.

  • 4
    $\begingroup$ Did you try integration by parts to get rid of the logarithm ? This would be simpler I guess. Continue with partial fraction decomposition. $\endgroup$ – Claude Leibovici Dec 13 '14 at 10:28
  • $\begingroup$ I considered it but then didn't proceed. I realise now that it may be the way to go. $\endgroup$ – rubik Dec 13 '14 at 10:30
  • $\begingroup$ Yes, it is ! Cheers :-) $\endgroup$ – Claude Leibovici Dec 13 '14 at 10:31

Let $x = \frac{1}{t}$

$$I = -\int \log(3 + 7t) - \log(t) dt$$

The rest is simple.

Your first thought was actually good.

  • $\begingroup$ Aw, I feel so dumb. I had forgotten logarithms' properties... $\endgroup$ – rubik Dec 13 '14 at 12:13
  • $\begingroup$ How do you go from the integral given in the post to your ? Substituting $x=\frac 1t$ does not seem to lead to your proposed form. But, for sure, I may be totally wrong. Cheers :-) $\endgroup$ – Claude Leibovici Dec 13 '14 at 13:46
  • $\begingroup$ The task is to simplify. After simplification, it does indeed lead to the proper answer. $\endgroup$ – Amad27 Dec 13 '14 at 14:17
  • $\begingroup$ This works, but you'd have to IBP for both of those log expression, so I think IBP at the beginning is easier $\endgroup$ – Dylan Dec 14 '14 at 9:36
  • $\begingroup$ Not really. We can simply use the integral table for $\log$ instead of IBP. $\endgroup$ – Amad27 Dec 14 '14 at 10:48

It turned out integration by parts was the way to go. I'll post here my solution in case someone needs it.

Recall the integration by parts formula: $$\int u\,\mathrm dv = uv - \int v\,\mathrm du$$

In this case $u = \ln(3x+7)$, and $\mathrm dv = 1/x^2\,\mathrm dx$.

Hence: $$\require{cancel} \begin{align}\int\frac 1{x^2}\ln(3x+7)\,\mathrm dx &= -\frac 1x\ln(3x+7) + \int\frac 1x\cdot\frac 3{3x + 7}\,\mathrm dx =\\ &=-\frac{\ln(3x+7)}x + 3\left(\frac 17\int\frac 1x\,\mathrm dx - \frac 1{\cancel{3}} \cdot \frac {\cancel{3}}7\int\frac{3}{3x + 7}\,\mathrm dx\right ) =\\ &= -\frac{\ln(3x+7)}x + \frac 37\ln x - \frac 37 \ln|3x + 7| + C =\\ &= \frac{-7\ln(3x + 7) + 3x\ln x - 3x\ln|3x + 7|}{7x} + C =\\ &= \frac{3x\ln x - (3x + 7)\ln(3x + 7)}{7x} + C \end{align}$$

  • $\begingroup$ You almost made it ! There is a sign error in the last line. Cheers :-) $\endgroup$ – Claude Leibovici Dec 13 '14 at 12:55
  • $\begingroup$ You are welcome ! $\endgroup$ – Claude Leibovici Dec 13 '14 at 16:14

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