Using the substitution rule to integrate $ \int_{-1}^{2} (1+4x^2)\,dx $ I am stuck on an integral problem that came out of an arc length problem.
I have an integral,
$$
\int_{-1}^{2}1+4x^2dx.
$$
When I try to apply the substitution rule, I am left with a variable that I don't know what to do with:
$$
u = 1+4x^2 \rightarrow du =8xdx \rightarrow dx = \frac{1}{8x}du.
$$
My question is as follows:
What do I do with the x in the denominator of dx...or is there a different way to solve this problem?
Thanks!
 A: If your primary objective is to find the arc length $L$ of the graph of the function $f(x)=x^2$ on the interval $-1\le x\le 2$, then the integral $\int_{-1}^{2}(1+4x^2)\,dx$ which you asked about in your question is not the integral you should be solving.
The correct integral for the arc length is:

$$L=\int_{a}^{b}\sqrt{1+\left[f^\prime{(x)}\right]^2}\,\mathrm{d}x\\
=\int_{-1}^{2}\sqrt{1+(2\,x)^2}\,\mathrm{d}x\\
=\color{blue}{\int_{-1}^{2}\sqrt{1+4\,x^2}\,\mathrm{d}x}.$$

Integrals of this type are most straightforwardly solved via trigonometric substitution. After substituting $x=\frac12\tan{(u)}$, the integral for $L$ becomes:
$$L=\frac12\int_{-\tan^{-1}{(2)}}^{\tan^{-1}{(4)}}\sec^3{(u)}\,\mathrm{d}u.$$
This is the famous "secant-cubed" integral. If you haven't encountered this integral before, you can read the "Integral of secant cubed" Wikipedia page which describes a variety of ways to solve this integral in detail. I leave the computation of the final answer as an exercise.
A: 1) please use MathJax,
2) you don't need substitution here at all: for the integrand $1+4 x^2$ just get two simple integrals. 
A: $$
\int_{-1}^{2}1+4x^2dx = \int_{-1}^2dx + 4\int_{-1}^2x^2dx = x|^{2}_{-1} + \frac{4}{3}x^3|^{2}_{-1} = (2-(-1))+\frac{4}{3}(2^3-(-1)^3) =15
$$
