Definite Integral of square root of polynomial I need to learn how to find the definite integral of  the square root of a polynomial such as: 
    $$\sqrt{36x + 1}$$ or $$\sqrt{2x^2 + 3x + 7} $$
EDIT: It's not guaranteed to be of the same form. It could be any polynomial that can't be easily factored into squares.
This isn't homework, I'm studying for a final. And for context, I'm finding the arc length of a function.

$$\int {\sqrt{36x+1}}\,dx$$
use this formula
$$\int {\sqrt{x}}\,dx=\frac {x^{\frac{3}{2}}}{\frac32}\implies \frac {2x^\frac32}{3}+C$$
so
$$\int {\sqrt{36x+1}}\,dx=\frac {2(36x+1)^\frac32}{(3)(36)}$$
$$\frac {(36x+1)^\frac32}{54}+C$$
Q2
$$\int\sqrt{2x^2 + 3x + 7}\,dx $$
$$\int\sqrt{2(x^2 + \frac{3x}{2} + \frac72)}\,dx $$
$$\int\sqrt{2(x^2 + \frac{3x}{2} + \frac72)}\,dx $$
$$\sqrt2\int\sqrt{(x^2 + \frac{3x}{2} + \frac72)}\,dx $$
$$\sqrt2\int\sqrt{(x^2 + \frac{3x}{2} + \frac72+\frac{3^2}{4^2}-\frac{3^2}{4^2})}\,dx $$
$$\sqrt2\int\sqrt{(x^2 + \frac{3x}{2} +\frac{3^2}{4^2}+ \frac72-\frac9{16})}\,dx $$
$$\sqrt2\int\sqrt{(x+ \frac{3}{4})^2 + \frac72-\frac9{16})}\,dx $$
$$\sqrt2\int\sqrt{(x+ \frac{3}{4})^2 + (\frac{\sqrt47}{4})^2}\,dx $$
now use this formula
$$\int{\sqrt{x^2+a^2}}\,dx=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2\log|x+\sqrt{x^2+a^2}|}{2}+C$$
I hope you can take it now on your own
