Question

is this right (integral problem)

Answer 1

\[\int\limits_{?}^{?} \frac{ 10 }{ (x^2+9)(x-1) }\]

Answer 2

i get A+C=0 -A+B=0 -B+9C=10

Answer 3

is that right?

Answer 4

i have Ax+B/x^2+9 + C/x-1

Answer 5

i dont know if im doing this right

Answer 6

I'm not sure how you got A+C=0 -A+B=0 -B+9C=10
take it from
\[\frac{Ax+B}{x^2+9} + \frac C{x-1}=\frac{10}{(x^2+9)(x-1)}\]and multiply both sides by \((x^2+9)(x-1)\)

Answer 7

i did that and i got Ax^2-Ax+Bx-B+Cx^2+9C

Answer 8

don't multiply it out, just leave the parentheses as is

Answer 9

why

Answer 10

you will see

Answer 11

it will make it much easier to find the constants

Answer 12

okay so i got (Ax+B)(x-1)+C(x^2+9) what do i do next

Answer 13

what is all that equal to?

Answer 14

10

Answer 15

ok write that
then let x=1

Answer 16

why 1?

Answer 17

plug it in and see what happens

Answer 18

i got c=1

Answer 19

correct, now the rest should be much easier to solve

Answer 20

now plug in C=1, then let x=0

Answer 21

i get B=-1

Answer 22

me too
now plug in whatever to get A

Answer 23

so the x values i can pick whatever number

Answer 24

yep

Answer 25

i got A=-1

Answer 26

me too, let's check wolfram...

Answer 27

yep, we seem to be right?

Answer 28

yeah so then i get -x-1/x^2+9 + 1/x-1 right?

Answer 29

yep

Answer 30

okay thank you i'll let you know if i have any more questions

Answer 31

i got -1/2x ln [x^2+] + ln[x-1] + C

Answer 32

looks good to me