Question

Help...Partial Fractions problems
1) x-20/x^2-4x; I have factored the denom.; set part the fractions using the rules, multiplied common denom., combined like terms... So it looks like this x-20/x(x-4)=A/x+ B/x-4; multiplied out it looks like this: x-20=A(x-4)+B(x); x-20=Ax-A4+Bx; (Ax+Bx)-A4; x(A+B)-A4...and then I'm lost...:(

Answer 1

You want to stop and evaluate the problem at this point here: x-20=A(x-4)+B(x)
What can you plug in for 'x' to get rid of each term individually?
If you let x=0, you get:
0-20 = A(0-4) + B(0) => -20 = -4A => A = 5

You got it from here. :)

Answer 2

After you solve for B, look back and recall that you have \(\int{}\frac{x-20}{x^{2}-4x}dx = \int{}(\frac{A}{x}+\frac{B}{x-4})dx\)
It becomes a matter of plug-n-play.

Answer 3

"x(A+B)-A4...and then I'm lost...:("

That's right. -4A must be equal to 20, and A+B must be equal to the coefficient of x.