Thread: Integration Question (inverse trig)

Question:
$\displaystyle \int \frac{dx}{x^2+6x+10}$

If it was $\displaystyle -6x$ I believe the method is to complete the square but not sure with $\displaystyle +6x$...

Would like some tips please

You would still complete the square.

$\displaystyle x^2+6x+10=(x+3)^2+1$, so we have

$\displaystyle \int \frac{dx}{(x+3)^2+1}=\int \frac{du}{u^2+1}$ (letting $\displaystyle u=x+3$).

Can you go from here?

Yah I got it, thanks a lot I have been feeling dumb a lot lately after some of my posts here...