I am asked to integrate this:
$\int_{0}^{2} [2\ln(2x+1)]dx$
So far I have solved it to this point:
$[[2x\ln(2x+1)]_{x=0}^{x=2} - \int_{0}^{2} (\frac{4x}{2x+1})]dx$
The answer is:
$[(2x\ln(2x+1)) - 2x + \ln(2x+1)]_{x=0}^{x=2}$
I do not understand how this second integral is performed. Possibly with partial fractions?
You have: $$\int_0^2 \frac{4x}{2x+1}\mathrm dx=2\int_0^2 \frac{2x}{2x+1}\mathrm dx=2\int_0^2 \frac{2x+1-1}{2x+1}\mathrm dx$$ Can you take it from here?
Instead of integration by parts I prefer sort of guessing the antiderivative. The first natural try is $2x\log(2x+1).$ However on differentiating we get $2\log(2x+1)$ and the additional term $2x{2\over 2x+1}=2-{2\over 2x+1}.$ Therefore we need to find the antiderivative of the last sum and subtract it from $2x\log(2x+1).$ The antiderivative of the sum is equal $2x-\log(2x+1).$ So the final result is $$\int 2\log(2x+1)\,dx =2x\log(2x+1)-2x+\log(2x+1)$$
Remark Similar procedure can be applied to the antiderivative of say $xe^{-x}.$ The first natural try is $-xe^{-x}.$ On differentiating we get $xe^{-x}$ and the additional term $-e^{-x}$ whose antiderivative is equal $e^{-x}.$ Therefore the final result is $-xe^{-x}-e^{-x}.$
I strongly support the substitution $u=2x+1$ suggested by @J. W. Tanner. In general, before starting any calculations it is worthwhile to simplify things as much as possible to make calculations simpler. Moreover, when expressions are simpler the chances of making mistakes are lower. Although in the OP example, when we make a substitution we have to remember about changing the interval of integration.
Substitution is another way to solve this integral. Let $u=2x+1$. Then $du=2dx$ and
$$\int_{0}^2 2\ln(2x+1)\,dx=\int_1^5\ln (u)\, du=\big(u\ln u-u\big)\,\bigg\rvert_{\,1}^{\,5}\;\;.$$
