Question

cmon guys, I really need help with this Calc II problem,
Evaluate the integral:

Answer 1

I would say check out theintegralcalc's channel on youtube. I'm crappy at integrals, but I think her channel will help.

Answer 2

mk, I'll check it out, I just got to finish up this drill first

Answer 3

Partial fractions

Answer 4

While you're at it, I'll search for a similar example.

Answer 5

I got this lol

Answer 6

http://www.youtube.com/user/TheIntegralCALC/search?query=integral+partial+fractions

Answer 7

(x+2)/[(x+7)(x-8)] = A/(x+7) + B/(x-8)

Answer 8

Multiplying across by (x+7)(x-8) you get:
A(x-8) + B(x+7) = x+2

Answer 9

Ax - 8A + Bx + 7B = x+2
(A+B)x +(-8A+7B) = x+2
A+B = 1
-8A+7B = 2

Answer 10

A = 1-B
-8(1-B)+7B = 2
-8 + 8B + 7B = 2
15B = 10, B = 10/15 = 2/3

Answer 11

B = 2/3
A = 1/3

Answer 12

(x+2)/[(x+7)(x-8)] = A/(x+7) + B/(x-8) = (1/3)/(x+7) + (2/3)/(x-8) =
= 1/[3(x+7)] + 2/[3(x-8)]

Answer 13

The third one.

Answer 14

There we go, now integrate that:
Int (1/[3(x+7)] + 2/[3(x-8)])
(1/3) Integral (1/(x+7) + 2/(x-8)) = (1/3) { Ln|x+7| + 2Ln|x-8| }

Answer 15

(1/3) { Ln|x+7| + 2Ln|x-8| } + C

Answer 16

2 ln ( (x-8) - ln ( x + 7)

- which is same as the third option

Answer 17

WOOT! No mistakes this time!

Answer 18

Wow, thanks guys! That really helps alot!

Answer 19

wait a second.. where did the 1/3 go?

Answer 20

first split the function into partial fractions then integrate each fraction

Answer 21

@bobdylan great question!

Answer 22

Hey as long as you guys are here, I can not seen to figure out what this problem is, or where to start, its a length of the curve one

Answer 23

I'd suggest making a new thread for your question and linking it.

Answer 24

b/c you might get more people in on it.

Answer 25

I don't even know where that f is coming from, ok

Answer 26

http://openstudy.com/study#/updates/4f52be85e4b019d0ebb059d7