Definite Integral of square root of polynomial
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 $$\int \sqrt{(x+b_1)^2 + c_1} dx = \dfrac{(b_1+x)\sqrt{P(x)} + c_1 \log \left(b_1 + x + \sqrt{P(x)}\right)}2 + \text{constant}$$
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.
Using the $u$-substitution $u = 36x + 1$, we have $du = 36dx$. Then, it follows that:
$$\int_{a}^{b} \sqrt{36x+1}dx = \frac{1}{36}\int_{36a+1}^{36b+1}u^{\frac{1}{2}}du = \frac{1}{36}*\frac{2}{3}u^{\frac{3}{2}}|^{36b+1}_{36a+1} = \frac{1}{54}(36x+1)^{\frac{3}{2}}|^{b}_{a}$$