Integrating $\int \frac{3}{x^2+12x+45}$ My professor is giving a mastery exam over integrals and one of the sections covered is un-factorable quadratics like this problem that are supposed to be done by completing the square.
$$\int \frac{3}{x^2+12x+45}$$
Completing the square and everything gives $$\int \frac{3}{(x+6)^2+9}$$
I figured the answer would be somewhere in the range of $$\frac{1}{3} \tan^{-1}\Biggl(\frac{x+6}{9}\Biggl)+c$$
like I learned in the textbook aside from not knowing what to do with the three. Anyways the answer is somehow $$\tan^{-1}\Biggl(\frac{x}{3}+2\Biggl) + c$$ and I don't know what I did wrong
 A: Since\begin{align}\int\frac3{x^2+12x+45}\,\mathrm dx&=\int\frac3{(x+6)^2+9}\,\mathrm dx\\&=\int\frac{\frac39}{\frac{(x+6)^2}9+1}\,\mathrm dx\\&=\int\frac{\frac13}{\left(\frac{x+6}3\right)^2+1}\,\mathrm dx,\end{align}the substitution $y=\frac{x+6}3$ leads indeed to the answer $\arctan\left(\frac{x+6}3\right)$.
A: I would suggest to use the formula
$$\int\frac{du}{u^2+a^2}=\frac{1}{a}\arctan\frac{u}{a}+C.$$
Now, we have $$\int\frac{3}{x^2+12x+45}dx=3\int\frac{dx}{(x+6)^2+9}.$$
With $a=3$ and $u=x+6$, you will get what you want.
A: You should have had $$\color{red}{3}\cdot\frac13\arctan\left(\frac{x+6}{\color{red}{3}}\right)+\text{some constant},$$ which indeed is $$\arctan{\left(\frac{x}{2} +2\right)}+\text{some constant}.$$
The correct formula is $$\int\frac{\mathrm d x}{(x+a)^2+b^2}=\frac1b\arctan\left(\frac{x+a}{b}\right)+\text{some constant}.$$
A: $$(x+6)^2+9=9\Bigl((\frac x3 +2)^2+1\Bigr)$$
put $$t=\frac x 3+2$$
with $dx=3dt$,
the integral becomes
$$\int \frac{3.3dt}{9(t^2+1)}=\arctan(t)+C$$
$$=\arctan(\frac x3+2)+C$$
