The integral $\int_0^8 \sqrt{x^4+4x^2}\,dx$ 
$\displaystyle \int\nolimits_0^8 \sqrt{x^4+4x^2}\,dx$.

Alright, so I thought I had this figured out. Here's what I did:


*

*I factor out an $x^2$ to get $\sqrt{x^2(x^2+4)}$.

*I let $x = 2\tan(\theta)$, therefore the integrand is $\sqrt{4\tan^2(\theta)  (4\tan^2(\theta) + 4)}$.

*Factor out a 4 and it becomes $\sqrt{(16\tan^2(\theta)  (\tan^2(\theta) + 1))}$

*Which equals $\sqrt{16\tan^2(\theta) \sec^2(\theta)}$

*This is easy to take the sqrt of. The integrand becomes $4\tan(\theta)\sec(\theta)$.

*Now, the integral of this is $4\sec(\theta)$

*And it's evaluated from $0$ to $\arctan(4)$ right? Because as $x$ goes to $0$, so does $\theta$, and as $x$ goes to $8$, $\theta$ goes to $\arctan(4)$...

*But the end result $(4 (\sec(\arctan(4)) - 1) )$ isn't the correct answer


I put it into WolframAlpha and I get $(8/3) (17\sqrt{17} - 1)$, which is the right answer. How did they get that? (there's no "show steps" option)
Any help is greatly appreciated!
PS, what's the syntax for doing sqrts and exponentials?

There are other ways of doing this integral, but let me try to fix your attempt, which is certainly a fine idea as far as it goes. 
The main problem I spot with your development is that you forgot to change the $dx$ when you did the change of variable. (And you should be able to evaluate $\sec(\arctan a)$ as well; we'll get to that shortly).
So: you start with
$$\int_0^8 \sqrt{x^4+4x^2}\,dx = \int_0^8 \sqrt{x^2(x^2+4)}\,dx.$$
Then you do the change of variable $x=2\tan(\theta)$. If you do this, then
$$dx = 2\sec^2\theta\,d\theta;$$
when $x=0$, you want $\theta=0$, and when $x=8$ you want $\theta=\arctan(4)$ (you are correct there). So the integral actually becomes, after changing integrand, limits, and the $dx$:
$$\begin{align*}
\int_0^8\sqrt{x^2(x^2+4)}\,dx &= \int_0^{\arctan(4)}\sqrt{4\tan^2\theta(4\tan^2\theta+4)}2\sec^2\theta\,d\theta\\
&= \int_0^{\arctan(4)} \sqrt{16\tan^2\theta(\tan^2\theta+1)}2\sec^2\theta\,d\theta\\
&= \int_0^{\arctan(4)}8\sec^2\theta\sqrt{\tan^2\theta\sec^2\theta}\,d\theta\\
&= 8\int_0^{\arctan(4)}\sec^2\theta|\tan\theta\sec\theta|\,d\theta.
\end{align*}$$
Now, on $[0,\arctan(4)]$, both tangent and secant are positive, so we can drop the absolute value signs (something else you were not careful with), and the integral becomes
$$8\int_0^{\arctan(4)}\sec^3\theta\tan\theta\,d\theta.$$
Set $u=\sec\theta$. Then $du=\sec\theta\tan\theta$, so we have
$$\begin{align*}
8\int_0^{\arctan(4)}\sec^3\theta\tan\theta\,d\theta &= 8\int_{\sec(0)}^{\sec(\arctan(4))}u^2\,du\\
&= \frac{8}{3}u^3\Biggm|_{\sec(0)}^{\sec(\arctan(4))}\\
&=\frac{8}{3}\left(\sec^3(\arctan(4)) - \sec^3(0)\right).
\end{align*}$$
Now, $\sec(0) = 1$. What about $\sec(\arctan(4))$?
Say $\psi$ is an angle with $\tan(\psi)=4$. Take a right triangle with this angle; by scaling, we may assume the opposite side has length $4$ and the adjacent side has length $1$. Then the hypotenuse has length $\sqrt{17}$, so the cosine of $\psi$ is $\frac{1}{\sqrt{17}}$, hence the secant has value $\sqrt{17}$. So $\sec(\arctan(4)) = \sec(\psi) = \sqrt{17}$. Thus, the integral is:
$$\begin{align*}
\int_0^8\sqrt{x^4+4x^2}\,dx &= \frac{8}{3}\left(\sec^3(\arctan(4)) - \sec^3(0)\right)\\
&=\frac{8}{3}\left( \sqrt{17}^3 - 1^3\right)\\
&= \frac{8}{3}\left(17\sqrt{17} - 1\right).
\end{align*}$$
In summary: your mistake was that when you did the change of variable, you forgot to change the differential as well; and at the end you could have simplified $\sec(\arctan(4))$. 
Of course, the better way of doing this is to factor out $x$ from the square root, and then recognize that you can do
$$\int_0^8\sqrt{x^4+4x^2}\,dx = \int_0^8x\sqrt{x^2+4}\,dx$$
with the change of variable $u=x^2+4$, like DJC suggested. But I thought you might like to know where exactly your approach went wrong (the substitution), and whether it could be brought to a correct conclusion (it could).
