Question about the integral of $1/(1+9x^2)$ This might be a stupid question but what is the integral of $\frac{1}{1 + 9x^2}$? I want to think it's $\tan^{-1}(3x) + c$ but the book I'm working in says $\frac{1}{3}\tan^{-1}(3x) + c$. 
How did they get $1/3$? 
 A: Another way to think about this question is to differentiate $\tan^{-1}(3x) + C$ and see what you get.  Using the chain rule, 
$$
\frac{d}{dx} \left ( \tan^{-1}(3x) + C \right ) = \frac{3}{1 + (3x)^2} = \frac{3}{1 + 9x^2}.
$$
Thus the correct antiderivative of 
$$\displaystyle \frac{1}{1+9x^2}
$$ 
must include a factor of $1/3$ to compensate: 
$$
\displaystyle \frac{1}{3} \tan^{-1}(3x) + C.
$$
A: Elaborating a bit more on the above hints in a more clear flow.
So we are trying to integrate the following expression $~~~\rightarrow ~~~ \displaystyle\int \dfrac{1}{1+9x^{2}}\ dx$. 
To do the this, we will need to make an appropriate substitution inside of the integrand and to make the $9x^{2}$ look like something equivalent. Doing this leads us to the following:
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\displaystyle\int \dfrac{1}{1+(3x)^{2}}\ dx$
Let: $~u=3x$
$du=3 dx$
$dx=\dfrac{1}{3}\ du$ 
Substituting in u and dx we see that we get the following:
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\displaystyle \dfrac{1}{3} \int \dfrac{1}{1+(u)^{2}}\ du$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=~\dfrac{1}{3} \text{arctan}(u)+C$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=~\dfrac{1}{3} \text{arctan}(3x)+C~~~~~~~~~~~~~~~~~~~~\blacksquare$
So as we can see, the $\dfrac{1}{3}$ came from the substitution we made and then going from there and solving for what dx represented in terms of u. 
Hope this helped out. Let me know if you have any questions on any of the steps made in this process. 
Thanks.
Good Luck.
A: Set $u=3x$. Then the integrand is $1/(1+u^2)$ which integrates to $\tan^{-1} u$ but $du=3dx$. This gives you the $1/3$.
