How to set up partial fractions using linear factors?

[philadelphia]

Homework Statement

I need to intergrate the following

(4x²+2x-1)/ (x³+x²)
How to set up problem using linear factors?

The Attempt at a Solution

Factoring the denominator I get:
x²(x+1)

By linear/quadratic factoring I get:
A/x + Bx+C/ x² + d/(x+1)

Is this right?

 

[Count Iblis]

Correct. All you need to do is to solve for the coefficients.

If you don't like to do that, you can use the following alternative method. We have:

R(x) =(4x^2+2x-1)/ (x^3+x^2) =

(4x^2+2x-1)/ [x^2(1+x)]

Near x = -1 this is:

R(x) = 1/(1+x) * {[4x^2 + 2 x - 1]/x^2 at x = -1} =

1/(1+x)

Now, a more accurate expansion of R(x) near x = 1 would be obtained by multiplying 1/(1+x) by the expansion of [4x^2 + 2 x - 1]/x^2 around the point x = 1, then you would get an expansion of the form:

1/(1+x) [1 + A (1+x) + B (1+x)^2 + ...] =

1/(1+x) + A + B(1+x)^2


But only the first term in ths expansion is singual, it diverges as you let x tend to 1.

If we now consider the expansion of R(x) near the other singular point, x = 0, and collect all the singular terms, we get:

(4x^2+2x-1)/ [x^2(1+x)] =

1/x^2 [4x^2 + 2x - 1] series expansion of 1/(1+x) =


1/x^2 [4 x^2 + 2 x - 1] [1-x + x^2 - ...]=

-1/x^2 + 3/x + nonsingular terms.

So, we have that near x = 1, the singular behaviour of R(x) is given by 1/(1+x) while near x = 0 it is given by -1/x^2 + 3/x. Then it follows that the function:

R(x) - [1/(1+x) - 1/x^2 + 3/x]

is a rational function that has no singularities because the only singular points of R(x) are at zero and at x = 1 but we have subtracted the singular terms of R around these points. It thus follws that:

R(x) - [1/(1+x) - 1/x^2 + 3/x]

is a polynomial. But since both R(x) and the subtracted terms tend to zero at infinity this polynomial must be identical to zero. Therefore we have:

R(x) - [1/(1+x) - 1/x^2 + 3/x] = 0 ---------->

R(x) = 1/(1+x) - 1/x^2 +3/x

And we didn't need to solve a single equation to get this result.