# 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 $1/3$?

-
Note that you integrate anything at wolframalpha (google it), then click show steps, and it will give you the step by step instructions for integrating! –  fdart17 Mar 17 '11 at 2:44
@Fdart: This is not particularly good advice. You didn't even answer --or even address in any way-- OP's question, which has a very straightforward answer if you just think about it rather than re-route yourself automatically to a computation tools... –  Bey Mar 17 '11 at 3:35
You can always check to see if your answer is correct by differentiation. If you differentiate $\tan^{-1}(3x)+C$, you'll get$$\frac{d}{dx}\left(\tan^{-1}(3x)+\right) = \left(\frac{1}{1+(3x)^2}\right)(3x)' = \frac{3}{1+9x^2}.$$So your answer is off by a factor of $3$. –  Arturo Magidin Mar 17 '11 at 3:56
@Fdart: in this case show steps in Alpha yields nothing. You get the integral, but no steps. But it is a reasonable thing to try first, and for OP to indicate that s/he OP has tried it and doesn't understand. –  Ross Millikan Mar 17 '11 at 4:12

Set $u=3x$. Then the integrand is $1/(1+u^2)$ which integrates to $\tan^{-1} u$ but $du=3dx$. This gives you the $1/3$.

-

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: $$\displaystyle \frac{1}{3} \tan^{-1}(3x) + C.$$

-

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.

-