Question

integral of sqrt (x^2+6x) dx

Answer 1

Try completing the square and then doing a trig substitution?

Answer 2

im not sure how to do the trig substitution..

Answer 3

i know the completing the square is (x+3)^2-3

Answer 4

Hmm I think it's (x+3)^2-9 right?
To complete the square you add and subtract 9.

Answer 5

\[\large \color{royalblue}{x^2+6x+9}-9 \qquad =\qquad \color{royalblue}{(x+3)^2}-9\]

Answer 6

Yeah, I dont know how to do this without trig sub unfortunately, lol.

Answer 7

well i dont know how to do it at all..

Answer 8

This calc 1 or 2?

Answer 9

2

Answer 10

Okay, makes my life easier. So trig sub is something you should be taught then. Okay, so the idea is if you ever have a form of either:
\[\sqrt{u^{2}-a^{2}}\]
\[\sqrt{a^{2}-u^{2}}\]or
\[\sqrt{u^{2}+a^{2}}\] where u is a variable or variable expression and a is a constant, then you can do a trig substitution.

when you have the form:
\[\sqrt{u^{2}-a^{2}}\], you let u = asec(theta) and the entire radical expression be atan(theta). So doing that, I can write this:

\[\int\limits_{}^{}3\tan \theta du\] The next step is to solve for du. Well, we just said u was asec(theta) or just 3sec(theta), meaning du = 3sec(theta)tan(theta)
giving us now:

\[9\int\limits_{}^{}\tan^{2} \theta \sec \theta d \theta\]do you think you would know how to integrate that? Now keep in mind, integrating this wont give you the finished answer, but would you know how to?

Answer 11

ill try. thanks

Answer 12

Yep. Once you integrate it, you have to go back to what our original substitutions were and use those to change the answer back into terms of x.