Arctangent integral How come this is correct: $$\int \dfrac{3}{(3x)^2 + 1} dx = \arctan (3x) + C$$
I learned that 
$$\int \dfrac{1}{x^2+1} = \arctan(x) + C$$
But I don't see how you can get the above one from the other. The $1$ in the denominator especially confuses me.

You can consider differentiation of composite function $(f \circ g)(x) = f(g(x))$ where $f(x) = arctan(x)$ and $g(x) = 3x$. We have
\begin{equation}
D(f(g(x))) = f'(g(x))g'(x) \ .
\end{equation}
Hence
\begin{equation}
\int f'(g(x))g'(x) dx = f(g(x)) + C \ .
\end{equation}
You may here want to recall that by definition $\int f(x) dx = F(x) + C$ iff $F'(x)=f(x)$. Now $f'(x) = \frac{1}{1+x^2}$ and $g'(x) = 3$ and we can calculate
\begin{eqnarray}
\int \frac{3}{1+(3x)^3} dx & = & \int f'(3x) 3 dx = \int f'(g(x)) g'(x) dx = f(g(x)) + C \\
& = & \arctan(3x) + C \ .
\end{eqnarray}
Note that in the first equation above we get rid of the confusing $1$ in the denominator.
