How do I solve a confusing trig substitution? Hi I am trying to solve an integral problem that involves trig substitution. First I tried completing the square, which gave me $1/\sqrt{(x+3)^2+2^2}$. I know I am supposed to use $x = \arctan(\theta)$. Does that mean it should be: 

$x + 3 = \arctan(\theta)$

$x = \arctan(\theta) - 3$

and then Integrate from there? I am not sure if this is a good way of thinking about this problem. I would appreciate any help!
So you are trying to find 
$$I=\int\frac{dx}{\sqrt{x^2+6x+13}}=\int\frac{dx}{\sqrt{(x+3)^2+2^2}}$$
And your goal is to find the right substitution $$x+3=a\cdot\tan\theta$$
such that $$(x+3)^2+2^2=(a\tan\theta)^2+2^2=2^2(\tan^2\theta+1)=2^2\sec^2\theta.$$
It does not take much to show that $a=2$. Can you take it from here?
Note that $$x^2+6x+13 = (x+3)^2+4$$

Your correct substitution  is$$(x+3)=2\tan(\theta)$$

$$dx=2\sec^2(\theta)d\theta$$

$$(x+3)^2+4=4\tan ^2(\theta)+4 =4(1+\tan^2 (\theta)= 4\sec^2(\theta)$$

$$\sqrt {x^2+6x+13}=2\sec (\theta)$$

Will take care of the radical and the integral is manageable from here on.