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.
On
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: $$\require{cancel} \begin{align}\int\frac 1{x^2}\ln(3x+7)\,\mathrm dx &= -\frac 1x\ln(3x+7) + \int\frac 1x\cdot\frac 3{3x + 7}\,\mathrm dx =\\ &=-\frac{\ln(3x+7)}x + 3\left(\frac 17\int\frac 1x\,\mathrm dx - \frac 1{\cancel{3}} \cdot \frac {\cancel{3}}7\int\frac{3}{3x + 7}\,\mathrm dx\right ) =\\ &= -\frac{\ln(3x+7)}x + \frac 37\ln x - \frac 37 \ln|3x + 7| + C =\\ &= \frac{-7\ln(3x + 7) + 3x\ln x - 3x\ln|3x + 7|}{7x} + C =\\ &= \frac{3x\ln x - (3x + 7)\ln(3x + 7)}{7x} + C \end{align}$$
Let $x = \frac{1}{t}$
$$I = -\int \log(3 + 7t) - \log(t) dt$$
The rest is simple.
Your first thought was actually good.