Calculus II - Trig Integration

Hey all, just wanted to confirm one answer and likely ask another question in a different thread;

$$\int\limits_{}^{} 1/(x^2+6x+12) dx $$
$$(x^2+6x+9) = (x+3)^2 + 3 $$
$$\int\limits_{}^{}1/a^2 + u^2 = (1/a)*\tan^-1(u/a) + C $$
$$1/\sqrt(3) * \tan^-1 ((x+3)/\sqrt(3)) + C $$

Can anyone confirm that answer?

Question

Calculus II - Trig Integration

Hey all, just wanted to confirm one answer and likely ask another question in a different thread; 

$$\int\limits_{}^{} 1/(x^2+6x+12) dx $$
$$(x^2+6x+9) = (x+3)^2 + 3 $$
$$\int\limits_{}^{}1/a^2 + u^2 = (1/a)*\tan^-1(u/a) + C $$
$$1/\sqrt(3) * \tan^-1 ((x+3)/\sqrt(3)) + C $$

Can anyone confirm that answer?

anonymous · Answer

ya

anonymous · Answer

Great, thanks! Now, I have another trig integration which I am actually just stuck on. Can I ask that here or should I post a new thread?

anonymous · Answer

Ill start typing it out here and probably post here and a new thread.

anonymous · Answer

Evaluate the integral; 

$$\int\limits(1/(x*\sqrt(x^8-9) dx$$
So I know this is in the form $$\int\limits du/u*\sqrt(u^2 - a^2) = (1/a) \sec^-1(|u/a|) + C$$

But I am not sure how to get that X and X^8 out of there. I know if the X was X^4 then X^8 would be (X^4)^2 but I am not really sure what to do. Any help starting this problem out would be great!