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Calculus/Integration problem

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Expert: - 4/27/2009


Question
Hello

My problem is integral of (x^3 + 5x^2 - 10x - 8)/(x^3 - 4x)

I'm pretty sure i'm supposed to start with long division, which i can figure out.

Then I think i factor the bottom into x(x^2 - 4) and then do something like (long division answer)/[x(x^2 - 4) = (A / x) + [B / (x^2 - 4)]

if this equation is correct, i can figure it out from this point. also, what is the long division answer i'm supposed to use? i thought i remembered only using the remainder or something? i could be wrong.

Thanks for all the help!

Answer
Divide it in once, so the power on top is less.
The result is (x^3 + 5x² - 10x - 8) - 1(x^3 - 4x) =
1 + (5x² - 6x - 8)/(x^3 - 4x).

In the first part, all you have to integrate is 1.
The result is x + C, where C is a constant.

Note that x^3 - 4x = x(x-2)(x+2).
So in the second part, it needs to be divided into
A/x + B/(x-2) + C/(x+2).

Combine the fractions and solve for A, B, and C.
You would have
[A(x²-4) + B(x²+2x) + C(x²-2X)]/(x^3-4x) = 5x² - 6x - 8.

From x², this would say that A+B+C = 5.
From x, this would say that 2B - 2C = -6.
From the constants, this would say that -4A = -8.

Once the constants are found, the integral of each of them is a ln().
For the first one, the function is A*ln(x), the second is B*ln(x-2), and the third is C*ln(x+2).

Also, remember the 1 that was out in front,
so add the integral of that in as well.

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