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Partial Fractions to find integral
- Thread starter kidmo87
- Start date
Hello, kidmo87!
We have two equal polynomials: .\(\displaystyle 3x - 4 \;=\;Ax + (B-A)\)
Equate corresponding coefficients.
. . \(\displaystyle \downarrow \text{- equate -} \downarrow\)
. . \(\displaystyle 3x - 4 \;=\;Ax + (B-A)\)
. . . . . . \(\displaystyle \uparrow \text{ -- equate -- }\uparrow\)
And we have: .\(\displaystyle \begin{Bmatrix}3 \:=\:A \\ \text 4 \:=\:A-B\end{Bmatrix}\)
\(\displaystyle \dfrac{3x-4}{(x-1)^2} \;=\;\dfrac{A}{x-1} + \dfrac{B}{(x-1)^2}\)
\(\displaystyle 3x-4 \;=\;A(x-1) + B\)
\(\displaystyle 3x - 4 \;=\;Ax - A + B\)
We have two equal polynomials: .\(\displaystyle 3x - 4 \;=\;Ax + (B-A)\)
Equate corresponding coefficients.
. . \(\displaystyle \downarrow \text{- equate -} \downarrow\)
. . \(\displaystyle 3x - 4 \;=\;Ax + (B-A)\)
. . . . . . \(\displaystyle \uparrow \text{ -- equate -- }\uparrow\)
And we have: .\(\displaystyle \begin{Bmatrix}3 \:=\:A \\ \text 4 \:=\:A-B\end{Bmatrix}\)
Last edited by a moderator:
D
Deleted member 4993
Guest
Hello everyone, could someone help me with understanding how A=3? thank youView attachment 2729
Multiply both sides by (x-1)2 to get:
3x - 4 = A*(x-1) + B
assign x = 1 then
3-4 = 0 + B → B = -1
then
3x - 4 = A*(x-1) + (-1)
3 *(x-1) = A*(x-1) → A = 3
Last edited by a moderator:
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,760
The important thing is that these equations have to be true for all x. So one thing you can do is put in specific values for x. For example, at the point where you have "3x- 4= A(x- 1)+ B" setting x= 1 gives 3- 4= A(1- 1)+ B or -1= B. Knowing that, Setting x= 0 gives 0- 4= A(0- 1)+ (-1) so that -4= -A- 1, -A= -3 and A= 3.