Question

I'm trying to use partial fractions to find the integral of 1/(3x-3x^2) dx ...

Answer 1

So what I have so far is:
I factored the denominator into (x)(-3x+3)
So
A/(x) + B/(-3x+3) = 1
so
A(-3x +3) + B(x) = 1

But now I don't know what to do ...

Answer 2

can we write
\[\int\limits_{?}^{?}\frac{ dx }{ 3x - x^2 } = \int\limits_{?}^{?}\frac{ dx }{ 3x(1-x) }\] ?

Answer 3

Sure, so then it would be
A/3x + B/(1-x)
so A(1-x) + B(3x) = 1
But same problem ... what now?

Answer 4

i will provide you easy solution
\[\frac{ 1 }{ 3 }\int\limits_{}^{}\frac{ 1 }{ x(1-x) }*dx\]
add and subtract x in the numerator
\[\frac{ 1 }{ 3 } \int\limits_{}^{} \frac{ x+1-x }{x(1-x) }*dx\]
do u follow these steps?

Answer 5

yes

Answer 6

now can u separate the terms?

Answer 7

what do u mean?

Answer 8

can we write above as
\[\frac{ 1 }{ 3 }\int\limits_{}^{} \frac{ x }{ x(1-x) } + \frac{ 1-x }{ x(1-x) }\] ?

Answer 9

OH. I see.

Answer 10

can u do the rest part?

Answer 11

i'm not really sure where you're going with this is the thing.

Answer 12

after cancelling the like terms, we will get
\[\frac{ 1 }{ 3 }\int\limits_{}^{} (\frac{ 1 }{ 1-x }+ \frac{ 1 }{ x })\]
i think the integration has left only
hope u can do this now

Answer 13

Yes! OK. I did my process differently but got this in the end as well.

Answer 14

if u want to solve the ques by your way then i can tell you that way too
i just told u different method

Answer 15

No, no, I get it. I like your method, also.

Answer 16

One more question: I'm getting 1/3 ln (p) + ln (1-p)
But I'm supposed to be getting 1/3 ln (p/1-p)
Where did I go wrong?

Answer 17

are u getting A= 1/3 and B = 1/3 ?

Answer 18

No, I'm getting A = 1/3 and B = 1

Answer 19

because -3A + B = 0 and A = 1/3 so B should be 1 right?

Answer 20

sorry my fault B should be 1

Answer 21

K

Answer 22

so u will be getting
\[\int\limits_{}^{}\frac{ A }{ x } + \frac{ B }{ (-3x+3) }\]

Answer 23

Yes

Answer 24

Use A=1/3 and B=1
and also make sure to take 3 out of the denominator

Answer 25

Right, I did use those values, which gives me 1/3 * (1/p + 1/(1-p))

Answer 26

and the integral of that is 1/3[ ln(p) + ln(1-p) ], right? But that's not the answer I'm supposed to be getting.

Answer 27

there would be "-ve" sign
u forgot to take the derivate of the terms inside natural log
{remember chain rule }

Answer 28

Oh. So 1/3 [ ln(p) - ln(1-p)?

Answer 29

yah now use the proprty of log

Answer 30

I'm supposed to get 1/3 [ ln (p/1-p) ], but I don't know why you divide by 1-p, that doesn't make sense.

Answer 31

OH! BECAUSE OF THE PROPERTIEES OF LOGS!

Answer 32

yeah
property of natural log is
1) lnA - lnB = ln(A/B)
2) lnA +lnB = ln(A*B)

Answer 33

YAY!!! OK thank you so much I understand this now :D

Answer 34

:-P