Integration of a second degree polynomial underneath a radical. I have a question about how to apply the substitution rule in an arc length problem. The problem asks for the arc length of $x^2$ on the interval $[-1,2]$.
Here is what I have come up with:
$$y=x^2
\frac {dy}{dx} = 2x
(\frac {dy}{dx}^2 = 4x^2$$
So, $$L = \int_{-1}^2 \sqrt {1+ \frac {dy}{dx}^2}dx
= \int_{-1}^2 \sqrt {1+4x^2}dx$$
So the question is, how do I solve the last step, i.e. $\int_{-1}^2 \sqrt {1+4x^2}dx$?
(If the mathjax is bad, please forgive me, I am just learning)

I shall give you the indefinite integral using IBP:
$$I=\int \sqrt{4x^2+1}\ dx=x\sqrt{4x^2+1}-\int \frac {(4x^2+1)-1}{\sqrt{4x^2+1}}\ dx$$
This means, $$2I=x\sqrt {4x^2+1}+\frac 12 \int \frac {dx}{\sqrt{x^2+(\frac 12)^2}}$$
Now, $$\int \frac {dx}{\sqrt {x^2+a^2}}=\int \frac {1+\frac {x}{\sqrt {x^2+a^2}}}{x+\sqrt {x^2+a^2}}\ dx=\ln|x+\sqrt {x^2+a^2}|+C$$
Hence we get:
$$I=\frac {x\sqrt {4x^2+1}}{2}+\frac 14 \ln|x+\sqrt {x^2+\frac 14}|+C$$
Now you can put the required limits to obtain the answer.
