Question

How would you calculate the integral of 3x/((x-a)(x-b))dx when a cannot be equal to b?

Answer 1

have you tried partial fractions

Answer 2

I did partial fraction and got A = -3a/b-a and B = -3b/a-b, but when it comes to integrating everything together I get -3a(ln(x-a))/b-a -3b(ln(x-b))/a-b. My online homework says it's wrong but i think I did everything correctly. Unless the formatting if off?

Answer 3

\[\text{ your work looks fine } \\ A \ln |x-a|+B \ln|x-b|+C \text{ where } A=\frac{3a}{b-a} \text{ and } B=\frac{-3b}{a-b}\]

Answer 4

like when you put in did you put parenthesis around the denominators?

Answer 5

3a*ln|x-a|/(b-a)-3b*ln|x-b|/(a-b)

Answer 6

Yes, I put parenthesis around the denominators, but even then I still keep getting it wrong.

Answer 7

We could try to write as one term
also it might also help to out the | | around the thing that is in the ln

Answer 8

I'll try changing the format of my answer and see if it works.

Answer 9

\[\frac{-3a}{a-b} \ln|x-a|+\frac{-3b}{a-b} \ln|x-b|+C \\ \frac{-3}{a-b}[a \ln |x-a|+b \ln|x-b|]+C \\ \frac{-3}{a-b}[ \ln|(x-a)^a|+\ln|(x-b)^b|+C \\ \frac{-3}{a-b}\ln|(x-a)^a(x-b)^b|+C\]

Answer 10

they it say anything special about how you should write it or simplify it

Answer 11

does it (not they)

Answer 12

It only says to calculate for a when it isn't equal to b. I keep getting it marked incorrectly.

Answer 13

can you show me how you enter it

Answer 14

and are you doing ln( ) or ln| |

Answer 15

It doesn't give me the option to do ln | |, it has to be ln ( ). My online homework is set up that way.

Answer 16

3a(ln(x-a))/(b-a) - 3b(ln(x-a))/(a-b)

Answer 17

and you did put the +C

Answer 18

yes

Answer 19

i notice when you wrote what you just wrote you put x-a in that second ln

Answer 20

if that isn't it i give up

Answer 21

should be 3a(ln(x-a))/(b-a) - 3b(ln(x-b))/(a-b)+C

Answer 22

Oh. Typo. But yea I just plugged it in again and nothing. Thank you so much for the help though. I'll see how to get around this mess.

Answer 23

\[\frac{A}{x-a}+\frac{B}{x-b}=\frac{3x}{(x-a)(x-b)} \\ A(x-b)+B(x-a)=3x \\ (A+B)x+(-Ab-Ba)=3x \\ A+B=3 \text{ and } -Ab-Ba=0 \\ A=3-B \\ -(3-B)b-Ba=0 \\ -3b+Bb-Ba=0 \\ B(b-a)=3b \\ B=\frac{3b}{b-a}=\frac{-3b}{a-b} \\ A=3-B=3+\frac{3b}{a-b}=\frac{3(a-b)+3b}{a-b}=\frac{3a}{a-b}\]

Answer 24

maybe we wrote A wrong

Answer 25

I just got a different A

Answer 26

Sorry for all the typos. I meant to put positive A.

Answer 27

But when it comes to integrating 3a/a-b * 1/x-a, you move 3a/a-b to the front of the integral correct?

Answer 28

\[\frac{3a}{a-b} \ln|x-a|-\frac{3b}{a-b} \ln|x-b|+C\]

Answer 29

It's still incorrect. Appreciate the help. I'm trying to see what the problem may be.

Answer 30

http://www.wolframalpha.com/input/?i=integrate%283x%2F%28%28x-a%29%28x-b%29%29%2Cx%29
this confirms my last answer there

Answer 31

i don't know what else could be the problem

Answer 32

I'm not sure either. But thank you!

Answer 33

sorry couldn't be more help
good luck