Evaluating $\int \ln(2x+3) \mathrm{d}x$

Question

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!

Answer 1

There is no mistake. $C$ is an arbitrary constant and $-\frac 3 2+C$ is just another constant $C'$. And there is no better way to answer this question.

Answer 2

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!

Answer 3

Alternative method

Consider, $$\frac{\mathrm{d}(x\ln(2x+3))}{\mathrm{d}x}=\ln(2x+3)+\frac{2x}{2x+3}$$ Rearranging, $$\ln(2x+3)=\frac{\mathrm{d}(x\ln(2x+3))}{\mathrm{d}x}-\frac{2x}{2x+3}$$ So intergrating both sides you can get the answer