Find A for Partial Fractions 1/(x+5)^2 (x-1)

[pillar]

1/(x+5)^2 (x-1)
B=(-1/6) C=(1/36)

I can't find the value of A, what method do you use to find it?

 

[rl.bhat]

1/(x+5)^2 (x-1)
B=(-1/6) C=(1/36)

I can't find the value of A, what method do you use to find it?

b]1/(x+5)^2 (x-1)[/b] = A/(x+5) + B/(x + 5)^2 + C/(x - 1)
Put x = 0.

 

[tanujkush]

1/(x+5)^2 (x-1)
B=(-1/6) C=(1/36)

I can't find the value of A, what method do you use to find it?

\frac{1}{(x+5)^{2}(x-1)} will disintegrate into \frac{Ax+B}{(x+5)^{2}} + \frac{C}{x-1}

Compare coefficients on both sides to get

A = -1/36
B = -11/36
C = 1/36

 

[pillar]

Ok thank, now what about this problem?

\frac{(x)^3}{(x+4)^{2}} will disintegrate into x+4=\frac{Ax+B}{(x+4)^{2}}

I'm not sure where to go from there, to get the values of A & B.

 

[tanujkush]

Ok thank, now what about this problem?

\frac{(x)^3}{(x+4)^{2}} will disintegrate into x+4=\frac{Ax+B}{(x+4)^{2}}

I'm not sure where to go from there, to get the values of A & B.

\frac{(x)^3}{(x+4)^{2}} will disintegrate into

\frac{(x+4-4)^3}{(x+4)^2}

which you can expand using the (a+b)^3[\tex\ standard formula and then its the same as the last one. Compare coefficients of powers of x on both sides to get A,B,C and so on.