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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 $\arcsin$?

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    $\begingroup$ 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. $\endgroup$ – John Omielan May 25 at 0:44
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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.

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

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