Integrals of Rational Functions

[Kp0684]

Integrals of Rational Functions...

The integral of:... (x-1)/x^4+6x^3+9x^2 , dx...i factored out the bottom getting: x^2(x+3)(x+3)...so, my new integral is: (x-1)/x^2(x+3)^2... now when i muiltlpy both sides by (x-1)/x^2(x+3)^2...i get... x-1= A(x+3)^2 + Bx^2(x+3) + C x^2...for A i got -1/9...and C= -4/9...B iam not getting an answer for that...A,B,C are coefficients once finding them you plug them back into the second equation...that will give the final answer...need help...

 

[Oggy]

You have to do it like this
[tex]\frac{x-1}{x^{2}(x+3)^{2}}=\frac{A}{x}+\frac{Bx+C}{x^{2}}+\frac{D}{x+3}+\frac{Ex+F}{(x+3)^{2}}[/tex]

 

[M98Ranger]

To solve for the coefficient B, you can use the method of partial fractions. First, you can rewrite the integral as:

(x-1)/x^2(x+3)^2 = A/x^2 + B/x + C/(x+3) + D/(x+3)^2

Then, you can multiply both sides by the denominator (x^2(x+3)^2) and equate the numerators:

x-1 = A(x+3)^2 + Bx^2(x+3) + Cx^2 + Dx

Next, you can substitute in different values for x to solve for the coefficients. For example, if you let x=0, you get:

-1 = A(3)^2 + C(0)^2

-1 = 9A

A = -1/9

Similarly, if you let x=-3, you get:

-4 = D(-3)

D = 4/3

You can solve for B and C by substituting in other values for x and solving for the coefficients. Once you have all the coefficients, you can plug them back into the original equation and integrate to get the final answer.