# Solving an exercise integral involving logarithms

I am asked to integrate this:

$$\int_{0}^{2} [2\ln(2x+1)]dx$$

So far I have solved it to this point:

$$[[2x\ln(2x+1)]_{x=0}^{x=2} - \int_{0}^{2} (\frac{4x}{2x+1})]dx$$

$$[(2x\ln(2x+1)) - 2x + \ln(2x+1)]_{x=0}^{x=2}$$

I do not understand how this second integral is performed. Possibly with partial fractions?

• Just long division $4x:2x+1$ Commented Jun 2 at 20:49
• Welcome to MSE. Try substituting $u=2x+1$ Commented Jun 2 at 20:50
• I actually never learned u substitution. My professor has a different way of teaching it, so u substitution is always confusing for me. Thank you though! Commented Jun 2 at 21:01
• Please, avoid chit-chating. A much better title would be something like "Help integrating $\int_{0}^{2} [2\ln(2x+1)]dx$". Commented Jun 2 at 21:24
• This question seems to be from at most Calculus 2. Commented Jun 2 at 23:06

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

$$\int_{0}^2 2\ln(2x+1)\,dx=\int_1^5\ln (u)\, du=\big(u\ln u-u\big)\,\bigg\rvert_{\,1}^{\,5}\;\;.$$