Weird integration issue: $\ln(x+1)=\ln(2x+2)$ ?!

Question

Weird integration issue: Using $(\ln[f(x)])'=\frac {f'(x)}{f(x)}$ we get that $\int \frac{2\,dx}{2x+2}=\ln(2x+2)$. Yet, $\int \frac{2\,dx}{2x+2}= \int\frac{dx}{x+1}=\ln(x+1)$ using the same rule as earlier.

What is wrong here?

Answer 1

$\ln(2x+2) = \ln 2 + \ln(x+1)$ (assuming $x > -1$). Antiderivatives are only determined up to an additive constant.

Answer 2

Nothing is wrong. You have found two different antiderivatives of the same function. You know that antiderivatives differ only by a constant on each interval where they are defined... you can check that this is the case in your computation.

Answer 3

Remember that:

$$\int \frac{2}{2x+2}dx \neq \ln(2x+2)$$

But:

$$\int \frac{2}{2x+2}dx = \ln(2x+2)+C$$

where $C$ is constant, so:

$$\int \frac{2}{2x+2}dx = \ln(2x+2)+C=\ln(2(x+1))+C=\ln(x+1)+\ln 2+C=\ln(x+1)+(\ln 2+C)=\ln(x+1)+C_1$$

So both calculations are almost correct, but you forgot about constant.

Answer 4

Just add a constant term and also notice that $\ln2$ is also a constant.