Partial Fractions Integral

[gadav478]

Hello all!

I need some help with this integral:

Evaluate the integral: int: [(x^3 + 3x^2 + x + 9)/(x^2 + 1)(x^2 + 3) ] dx

using the linear quotient rule I have:
let x^3 + 3x^2 + x + 9 = Ax+B (x^2 + 3) + Cx + D(x^2 + 1)
I am trying to solve for A B C and D...

I am unsure how to go about solving for them. The book says to equate the corresponding coefficients but I am not sure which. Are they corresponding with respect to the degree or the term number?

Any help is appreciated! Thanks!

 

[Deleted member 4993]

Hello all!

I need some help with this integral:

Evaluate the integral: int: [(x^3 + 3x^2 + x + 9)/(x^2 + 1)(x^2 + 3) ] dx

using the linear quotient rule I have:
let x^3 + 3x^2 + x + 9 = (Ax+B) (x^2 + 3) + (Cx + D)(x^2 + 1)

Those parentheses are essential!

I am trying to solve for A B C and D...

I am unsure how to go about solving for them. The book says to equate the corresponding coefficients but I am not sure which. Are they corresponding with respect to the degree or the term number?

Any help is appreciated! Thanks!

x^3 + 3x^2 + x + 9 = (Ax+B) (x^2 + 3) + (Cx + D)(x^2 + 1)

x^3 + 3x^2 + x + 9 = (Ax^3+ 3Ax + Bx^2 + 3B) + (Cx^3 + Cx + Dx^2 + D)

x^3 + 3x^2 + x + 9 = (A+C)x^3+ (3A+C)x +(B+D)x^2 + (3B+D)

Equating coefficients of same powers of x:

A + C = 1

3A + C = 1

B + D = 3

3B + D = 9

You have four equations and four unknowns - solve for A, B, C & D

 

[gadav478]

Thank you so much. I was confused because I wasn't treating Ax+ B as a factor entirely. Now I have a coefficient for x^3 on both sides. Thanks a lot.

 

[lookagain]

I need some help with this integral:

Evaluate the integral: int: [(x^3 + 3x^2 + x + 9)/((x^2 + 1)(x^2 + 3))] dx

\(\displaystyle \ \ \ \ \ \)If you are meaning \(\displaystyle \ \int\dfrac{x^3 + 3x^2 + x + 9}{(x^2 + 1)(x^2 + 3)}dx, \ \ \) then you need the additional grouping symbols, as suggested in red above.