Find $\int \frac{1}{\sqrt{-x^2-6x+40}}dx$ using completing the square?

Question

I am not sure how to find the integral by completing the square here since it's inside of a square root.

I am practicing with Khan Academy, and I have four choices for answers, all of which include either $\arcsin$ or $\arctan$.

$$\int \frac{1}{\sqrt{-x^2-6x+40}}dx$$

$$=\int \frac{1}{\sqrt{-(x^2+6x-40)}}dx$$

$$=\int \frac{1}{\sqrt{-(x^2+6x+9-9-40)}}dx$$

$$=\int\frac{1}{\sqrt{-(x+3)^2-7^2}}dx$$

I am not sure how to go farther than this. How can I get this to resemble the derivative of $\arctan$ or $\arcsin$?

Answer 1

Hint: It should be $+7^2$ instead. Once you have that, your integral will look like $$\int\frac{1}{\sqrt{a^2-u^2}}\, du,$$ which you should recognise.

Answer 2

You should get $+7^{2}$ under the square root. The substitution $x+3 =7 sin \theta$ gives the answer easily.