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

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 $$\arcsin$$?

• In your last line, you forgot to use the negative of the outside of the brackets with the negative inside. You got $-7^2$ when it should be $7^2$ instead. – John Omielan May 25 '19 at 0:44

## 2 Answers

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.

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