integral (x^3)/(x^2-2x+1) from -1 to 0
How do you get this integral? I was doing partial fractions and got A/(x-1)+B/(x-1), A+B=x^3, but that should equal 0 so where do I go from there?
Since (x-1) is a repeated factor, you need A/(x-1) + (Bx+C)/(x-1)^2
First, a little long division will yield
x+2 + (3x-2)/(x-1)^2
That further breaks down into
x + 2 + 3/(x-1) + 1/(x-1)^2
Now it's clear sailing.
For some reason I had completely overlooked the Bx+C and long division part. Thanks though!
To solve the integral of (x^3)/(x^2 - 2x + 1) from -1 to 0, partial fractions is indeed the right approach.
You correctly started with the partial fraction decomposition:
(x^3)/(x^2 - 2x + 1) = A/(x - 1) + B/(x - 1)
To proceed, we need to find the values of A and B.
Since the denominators are the same, we can equate the numerators:
x^3 = A + B(x - 1)
Before we move forward, let's expand the right side:
x^3 = A + Bx - B
To isolate A and B, let's compare the coefficients of x^3, x^2, x, and the constant terms on both sides of the equation.
Comparing the coefficients of x^3:
1 = 0
This implies that A = 0.
Now, comparing the coefficients of x^2:
0 = 0
This doesn't give us any additional information.
Next, comparing the coefficients of x:
0 = B
We have found that B must be 0.
Now, we can rewrite our partial fraction decomposition:
(x^3)/(x^2 - 2x + 1) = 0/(x - 1)
Since B = 0, the integral becomes:
∫(0/(x - 1)) dx
Integrating 0 gives us 0 as the result:
0∫ dx = 0
Therefore, the value of the integral is 0.
In summary, the computed integral is 0.