integate $\int\sqrt{x^2+3x+3} dx$ Is there any universal method how to solve integrals like this?
$$\int\sqrt{x^2+3x+3} dx$$
Or this?
$$\int\sqrt{-x^2+3x+3} dx$$
I tried use first Euler subs, but it was not good idea. 
$$\sqrt{ax^2+bx+c} = t\pm\sqrt{ax}$$
 A: For the first: complete the square to rewrite this as
$$
\sqrt{(x+3/2)^2 + 3/4}
$$
Now, apply a substitution $\theta$ such that
$$
x+3/2= \sqrt{3/4}\tan \theta
$$
For the second: complete the square to rewrite the integrand as
$$
\sqrt{k - (x-3/2)^2}
$$
For some $k$ that I'm too lazy to calculate. Then, apply a substitution such that
$$
x-3/2= \sqrt k \sin \theta
$$
A: For any integral of the forms, you have mentioned, 
transform the integrand into the sum or difference of two perfect square expressions and the resulting expression will look like  $\sqrt{(x\pm a)^2+(b)^2}$ ,$\sqrt{(x\pm a)^2-(b)^2}$  or $\sqrt{(b)^2-(x\pm a)^2}$.
And then use the following formulae which can be deduced with the help of integration by parts:


*

*$$\sqrt{(x\pm a)^2+(b)^2}dx=\frac{(x\pm a)\sqrt{(x\pm a)^2+(b)^2}}{2}+\frac{b^2}{2}\ln |(x\pm a)+\sqrt{(x\pm a)^2+(b)^2}|$$

*$$\sqrt{(x\pm a)^2-(b)^2}dx=\frac{(x\pm a)\sqrt{(x\pm a)^2-(b)^2}}{2}-\frac{b^2}{2}\ln |(x\pm a)+\sqrt{(x\pm a)^2-(b)^2}|$$

*$$\sqrt{(b)^2-(x\pm a)^2}dx=\frac{(x\pm a)\sqrt{(b)^2-(x\pm a)^2}}{2}+\frac{b^2}{2}\sin \frac{x\pm a}{b}$$


Hope you can apply this to your problem and solve it now.
A: By a linear transform of the argument, you can normalize the integrand to one of
$$\sqrt{1+x^2}\text{ or }\sqrt{1-x^2}.$$
Then by parts,
$$\int\sqrt{1\pm x^2}\,dx=x\sqrt{1\pm x^2}\mp\int\frac{x^2}{\sqrt{1\pm x^2}}dx=x\sqrt{1\pm x^2}-\int\frac{(1\pm x^2)-1}{\sqrt{1\pm x^2}}dx.$$
Move the first part of the integral to the LHS and remains to integrate
$$\int\frac{dx}{\sqrt{1\pm x^2}},$$ which is elementary.
