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.
 A: $$\int {\sqrt{36x+1}}\,dx$$
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
A: For a linear polynomial under the radical a $u$-substitution will do. If you wish to solve this problem for higher order polynomials in the general case then you will be appealing to a lot of algebra, hyperbolic trig functions, and elliptic integrals (which are similar in that they are arclength problems). In short there is no easy short cut to solve this problem for arbitrary degree polynomials.
For degree one polynomials, do as the other answers have advised. For degree two polynomials, like $\sqrt{2x^{2}+3x+7}$, you should employ trigonometric substitutions.
Step 1: Given $\int\sqrt{ax^{2}+bx+c}$ first complete the square into something of the form $k\int\sqrt{\pm u^{2}\pm l}$.
Step 2: Using one of the three substitutions from this article on the matter to get the integral into the form of $k\int\sqrt{f(x)^{2}}$ for some trig function $f(x)$.
Step 3: Remove the radical and solve using known integrals. 
If you wish to venture into solving cubics (or higher degree polys) in the general case (i.e. not reducible by a substitution) then do read up on elliptic integrals.
A: Using the $u$-substitution $u = 36x + 1$, we have $du = 36dx$. Then, it follows that:
$$\int_{a}^{b} \sqrt{36x+1}dx = \frac{1}{36}\int_{36a+1}^{36b+1}u^{\frac{1}{2}}du = \frac{1}{36}*\frac{2}{3}u^{\frac{3}{2}}|^{36b+1}_{36a+1} = \frac{1}{54}(36x+1)^{\frac{3}{2}}|^{b}_{a}$$
A: Recall that for $\alpha \neq -1$, we have
$$\int (ax+b)^{\alpha}dx = \dfrac1a \cdot \dfrac{(ax+b)^{\alpha+1}}{\alpha+1} + \text{ constant}$$
A way to see the above is as follows. Let $y = ax+b$. We then have $dy = adx$. Hence for $\alpha \neq -1$,
$$\int (ax+b)^{\alpha}dx = \int y^{\alpha} \dfrac{dy}a = \dfrac1a \dfrac{y^{\alpha+1}}{\alpha+1} + \text{ constant} = \dfrac1a \dfrac{(ax+b)^{\alpha+1}}{\alpha+1} + \text{ constant}$$
If $\alpha = -1$, we then have
$$\int \dfrac{dx}{ax+b} = \dfrac{\log(ax+b)}a + \text{ constant}$$

In general, there is no easy way to get $$\int \sqrt{P(x)} dx,$$ if degree of $P(x)$ is greater than $2$.
If $P(x)$ is linear, i.e., has degree $1$, I have mentioned above how to proceed.
Below we will see how to proceed if $P(x)$ is quadratic, i.e., $$P(x) = ax^2 + bx + c = a ((x+b_1)^2 + c_1).$$
$$b_1=\frac{a}{2b}$$ $$c_1=\frac{c}{a}-b_1^2$$
We now have the following
$$\sqrt{P(x)} = \sqrt{a} \sqrt{(x+b_1)^2 \pm c_1}$$
which gives us that
$$\int \sqrt{P(x)} dx = \sqrt{a} \int \sqrt{(x+b_1)^2 + c_1} dx$$
and
$$\int \sqrt{(x+b_1)^2 + c_1} dx = \dfrac{(b_1+x)\sqrt{P(x)} + c_1 \log \left(b_1 + x + \sqrt{P(x)}\right)}2 + \text{constant}$$
