Question

\[\int \frac{1}{\sqrt{y^{2}+8y+15}}dy\]

Answer 1

There are two ideas I have, you can either factor it which seems obvious, or complete the square which is what ends up being useful here.

Answer 2

=/ i hate completing the square.. I think I've mentally block it out haha can't even remember how to do it

Answer 3

isn't it something like
\[y^{2}+8y+15=0\rightarrow y^{2}+8y+()=15\]

Answer 4

Hahaha fair enough, I think everyone feels that way about it to be honest.

I think it's easier if you sorta think 'backwards' and start from here.
\[(y+a)^2 =y^2+2ay+a^2\]

Answer 5

\[y^{2}+8y+16=15+16\rightarrow (y+4)^{2}=31\]

Answer 6

ok that makes sense.. from what i did what do i do with the 31?

Answer 7

Almost, I think there's gotta be some kinda mistake, at the end of the day what you get has to be equal to the original thing, \(y^2+8y+15\)

Answer 8

I think you can figure out your mistake if I show you the answer:

\[(y+4)^2-1=( y^2+8y+16)-1\]

Answer 9

right, i get it your way I think i was trying to solve for y or something... anyways on to the int

Answer 10

Yeah all good, I think the integral is now of a form on your chart, so shouldn't be too crazy.

Answer 11

\[\int \frac{1}{\sqrt{(y+4)^{2}-1}}dy\rightarrow ln|y+4+\sqrt{(y+4)^{2}-1}|+C\]

Answer 12

Yeah that's the stuff good job.

Answer 13

thank you