<s>[INST] Evaluating $\int \ln(2x+3) \mathrm{d}x$ Evaluate $$\int \ln(2x+3)\mathrm{d}x$$
Set $r = 2x+3 \rightarrow \mathrm{d}r = 2\mathrm{d}x$
So integral becomes $\displaystyle \frac{1}{2}\int \ln(r)\mathrm{d}r$
Set $u=\ln(r)$, $\mathrm{d}v=\mathrm{d}r$, so $\mathrm{d}u = \frac{1}{r}\mathrm{d}r$ and $v=r$
$\implies \displaystyle \frac{1}{2}\int \ln(r)\mathrm{d}r=\frac{1}{2}\int u\, \mathrm{d}v=\frac{1}{2}(uv-\int v\, \mathrm{d}u) = \frac{1}{2}\left(\ln(r)r-\int \mathrm{d}r\right) = \ln(2x+3)\left(x+\frac{3}{2}\right)-\left(x+\frac{3}{2}\right)+C$
But the correct answer is $\ln(2x+3)\left(x+\frac{3}{2}\right)-x+C$
Can somebody show me where my mistake is and also a better way to do the problem? Thanks!
 [/INST] Your solution is correct as a constant plus another constant can be represented by a different constant, so $-\frac{3}{2}+C=C_1$.
As an alternative, you could integrate by parts and let $u=\ln(2x+3)$ and $dv=dx$. Then $du=\frac{2}{2x+3}$ and we can take $v=x+\frac{3}{2}$. It follows that
\begin{align}\int \ln(2x+3)\,dx&=\ln(2x+3)\left(x+\frac{3}{2}\right)-\int \left(x+\frac{3}{2}\right)\frac{2}{2x+3}\,dx\\&=
\ln(2x+3)\left(x+\frac{3}{2}\right)-\int \,dx\\&=
\ln(2x+3)\left(x+\frac{3}{2}\right)-x+C,
\end{align}
as expected!
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