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.

We can say even more in the general case: if a function $\;f\;$ is derivable , then
$$\int \frac{f'(x)}{1+f(x)^2}dx=\arctan(f(x)) + K(=\;\text{a constant})$$
which you can quickly verify by differentiating applying the chain rule.
In your particular case we simply have $\;f(x)=3x\;$ ...
