Calculus contradiction? 
Why is $$ \int \dfrac{1}{2x+2}dx \neq \dfrac{1}{2}\int \dfrac{1}{x+1} dx?$$

According to WolframAlpha: http://www.wolframalpha.com/input/?i=integral+of+1%2F%282x%2B2%29+%3D+1%2F2%28integral+of++1%2F%28x%2B1%29%29
They should be equal but why are they not?
 A: Wolfram Alpha left out the constant of integration:
$$
\begin{aligned}
\int \frac{1}{2x+2} \, dx
&= \frac{1}{2}\log(2x + 2) + C \\
\frac{1}{2} \int \frac{1}{x+1} \, dx
&= \frac{1}{2}\log(x + 1) + C^\prime.
\end{aligned}
$$
But
$$
\frac{1}{2}\log(2x + 2)
= \frac{1}{2}\big(\log(x + 1) + \log(2)\big)
= \frac{1}{2}\log(x + 1) + \frac{\log(2))}{2}.
$$
That is, the two antiderivatives differ by a constant.
A: The problem here is indeed the constant. When you evaluate a indefinite integral
$$\int f(x)dx$$
the result is a set of functions $F(x)+C$ where $F'(x)=f(x)$ and $C\in \Bbb R$. It appears that wolfram by default when you write "integral of something" inside an expression, such as your "=", decides to forget about this set and instead takes a single representative. In your case
$$\frac{1}{2}\log (2x+2)=\frac{1}{2}\log (x+1)+\frac{\log2}{2}$$
A: 1)
$$\int\frac{1}{2x+2}\space\text{d}x=$$

For the integrand $\frac{1}{2x+2}$, substitute $u=2x+2$ and $\text{d}u=2\space\text{d}x$:

$$\frac{1}{2}\int\frac{1}{u}\space\text{d}u=\frac{\ln\left|u\right|}{2}+\text{C}=\frac{\ln\left|2x+2\right|}{2}+\text{C}$$
2)
$$\int\frac{1}{x+1}\space\text{d}x=$$

For the integrand $\frac{1}{x+1}$, substitute $u=x+1$ and $\text{d}u=\space\text{d}x$:

$$\int\frac{1}{u}\space\text{d}u=\ln\left|u\right|+\text{C}=\ln\left|x+1\right|+\text{C}$$
