how to integrate this fucntion please?

Question

This is part of a differential question but i am stuck at this integration part. any suggestions i will really appriciate. ty.

intergrate 1/(a - x)(b - x) with repect to X.

the whole question is basically solve for x

dx/dt = k(a - x)(b - x)
first I separte the varibles which gives

dx / (a - x)(b - x) = k dt

Answer 1

You can try splitting 1/[(a- x)(b- x)] into two terms, namely
___c/(a- x)+ d/(b- x)

Let the original fraction equal the new sum of fractions,
___c/(a- x)+ d/(b- x) = [c(b- x)+ d(a- x)]/[(a- x)(b- x)] =
___= 1/[(a- x)(b- x)]

Thus,
___[c(b- x)+ d(a- x)] = 1

Set what has x's apart from what doesn't, and solve for c and d in terms of a and b.
___cb- cx+ da- dx = 1
___(cb+ da)- x(c+ d) = 1

Notice that there are no x's to the right-hand side, which means that
___c+ d = 0.....[01]
and
___cb+ da = 1 [02]

Rearranging [01], you get
___d = -c...... [03]

Replacing d for -c at [02], you get
___-db+ da = 1
___d(a- b) = 1
___d = 1/(a- b)

Thus,
___c = 1/(b- a)

Which means that
___1/[(a- x)(b- x)] = 1/[(b- a)(a- x)]+ 1/[(a- b)(b- x)]

The integral then becomes
___Integral{ dx/[(a- x)(b- x)] } =
___= Integral{ dx/[(b- a)(a- x)]+ dx/[(a- b)(b- x)] }
___= 1/(b- a)*Integral{ dx/(a- x) } + 1/(a- b)*Integral{ dx/(b- x) }
___= 1/(a- b)*ln(a- x)+ 1/(b- a)*ln(b- x)+ C

Answer 2

Sorry, Heinz, I just HAD to see if I could still do it!

Using the Heaviside expansion,

k/((a-x)*(b-x)) == k/((x-a)*(x-b) == A/(x-a) +B/(x-b)

k/(x-b) == A + B*(x-a)/(x-b), A=k/(a-b)

k/(x-a) == A*(x-b)/(x-a) + B, B=k/(b-a)

S(k*dx/((a-x)*(b-x)) = (k/(a-b))*S(dx/(x-a) + (k/(b-a))*S(dx/(x-b)

...Where S is substituted for the German S because I'm too lazy to figure out how to set up symbols on my keyboard...

After the integration getting the desired form x=f(t) is a matter of algebraic manipulation. In fact, the whole problem is pretty much an exercise in algebraic manipulation.

Answer 3

I shall not solve it but tell the steps

let 1/(a-x)(b-x) be A/(a-x) + B(b-x)
then 1 = A(b-x) + B(a-x)
solve these for A and B

now integrate each part

1/(a-x) = -1/(x-a) and integral is -ln(x-a)

Answer 4

Multply the factors to get

1/(x^2 -(a+b)x+ab).

You can find this integral in a table (since you are trying to solve a differntial equation, you are not learning integrals, so I think this is valid. The integral dx/(Ax^2 +Bx+C) is

2/sqrt(4AC-B^2) * arctan[2Ax+B)/sqrt(4AC-B^2)] for 4AC-B^2 > 0
2/sqrt(B^2-4AC) * arctanh[(2Ax+B)/sqrt(B^2-4AC for 4AC-B^2 < 0

In your problem A=1 B=-(a+b) and C=ab.

Answer 5

To get free school work help:

The Discovery Channel offers Cosmeo.com, while AOL has StudyBuddy.com. Then there is HomeworkSpot.com along with Ask for Kids (www.askforkids.com). Also, NationalGeographic.com/homework, SparkNotes.com, FigureThis.org and Tutor.com and Brainfuse.com

Answer 6

make the bottom (a-x)(b-x) into -1 exponents and move them to the top

i find fractions hard to work with...

once you move them up, expand it and take the integral... i hope that helps a bit

plus... at the texas instruments website, they have a program u can download for free to solve the integral of equations right on your calculator at teh push of a button

*this is how i'm surviving bc calculus ap

Answer 7

The trick to doing this type of problem is to separate it into 2 integrals by realizing:

1/(a-x) - 1/(b-x) = (b-a)/(a-x)(b-x)

so you could rewrite your whole question like:

1/(b-a) * S 1/(a-x) - S 1/(b-x)

and then just solve it like you normally would:
1/(b-a) * (-ln|a-x| - -ln(b-x))
1/(b-a) * (ln|1/(a-x)| + ln|b-x|)
1/(b-a) * ln|(b-x)/(a-x)|

also possible to write this answer as
1/(a-b) * ln|(a-x)/(b-x)|

Enjoy

Answer 8

1/(ab-ax-bx-^2x)
=1/(ab-ax-bx-^2x)^2{-2(x+1)}x
dx/dt
where dx/dt= k(a-x)(b-x)

Please don't appriciate in the public.

Answer 9

1/(a-x)(b-x) = 1/(a-b) * (a-b)/(a-x)(b-x)

= 1/(a-b) * (a-x+x -b)/(a-x)(b-x)

= 1/(a-b) * [ (a-x) -(b-x) ] / (a-x)(b-x)

= 1/(a-b) * [ (a-x) /(a-x)(b-x) - (b-x)/(a-x)(b-x)]

= 1/(a-b) * [ 1/ (b-x) -1/ (a-x) ]

NOW INTEGRATING = 1/ (a-b) * [log (b-x) - log (a-x)]*(-1)

= 1/(a-b) * log[ (b-x)/(a-x) *(-1)

= 1/(a-b) * log [ (a-x)/(b-x)] +A.constant