Integral of 1/(8+2x^2)

I have been following a rule saying that $$\int{\frac{1}{a^2+u^2}}dx = \frac{1}{a}\tan^{-1}(\frac{u}{a})+c$$

The question is asking for the interval of $$\frac{1}{8+2x^2}$$

Following that rule

$$a=\sqrt8$$ $$u=2x$$

So

$$\int{\frac{1}{8+2x^2}}dx = \frac{1}{\sqrt8}\tan^{-1}(\frac{2x}{\sqrt{8}})+ c$$

This seemed to have worked in the past but Wolfram is saying it is equal to

$$\frac{1}{4}\tan^{-1}(\frac{x}{2})+ c$$

They have used the rule stating $$\int{\frac{1}{u^2+1}}dx=\tan^{-1}(u)+c$$

And they factor out the constants of the equation to get to that form.

My question is, is the way I am doing it okay, or should I be adopting the other method?

-
Your problem is that you set $u = 2x$, but $2x^2 \neq (2x)^2$. –  Daniel Fischer Apr 9 '14 at 9:02
–  Double AA Apr 9 '14 at 13:26

You can do the change of variable, but use $u=\sqrt{2}x$. And remember that $du=\sqrt{2}dx$.

I think the best way is to factorize out a 2 to get $$\frac{1}{2}\int\frac{1}{4+x^2} \, dx$$ and then use the rule you stated.

-

$$\int\frac{dx}{8+2x^2}=\frac18\int \frac{dx}{1+\left(\frac{x}{2}\right)^2}=\frac14\int \frac{d\left(\frac x2\right)}{1+\left(\frac{x}{2}\right)^2}=\frac14\arctan\left(\frac x2\right)+C$$

-

Both rules yield the same result if do things properly.

In this case you forgot to change $du = 2dx$; Write explicitly the integral after the change of variable, without forgetting to change the differential

-