Integrate f(x) = (x^3 + 3x + 12) / (x(x+2)^2)

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In summary, the conversation discusses the integration of the function f(x) = (x^3 + 3x + 12) / (x(x+2)^2) and suggests rewriting it as a sum of partial fractions in order to solve it correctly. The process of solving a system of equations to find constants A, B, and C is also mentioned as a possible method.
  • #1
Hevonen
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Homework Statement



Integrate f(x) = (x^3 + 3x + 12) / (x(x+2)^2)

Homework Equations




The Attempt at a Solution



I know that i should somehow rewrite the polynomial but I do not know how. Please, help me.
 
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  • #2
Hmm, I don't know if this is correct or not, but have you tried long division?
 
  • #3
You should rewrite your polynomial as sum of partial fractions, hope you know how to do that. You write
[tex]f(x) = \frac{x^3 + 3x + 12} { x(x+2)^2} = \frac{A}{x}+\frac{B}{x+2}+\frac{C}{(x+2)^2} [/tex]

Where A,B,C are constants. Then you have to solve a system of three equations for A,B,C. Else you look in math book or wikipedia under partial fractions.
 
  • #4
Thanks! It is about partial fractions.
 
  • #5
I would strongly recommend that you multiply out the denominator and divide first. "Partial fractions" only works correctly when the numerator is of lower degree than the denominator.
 

1. What is the domain of the function?

The domain of the function is all real numbers except x = 0 and x = -2, since these values would result in division by zero.

2. What is the range of the function?

The range of the function is all real numbers, since there are no restrictions on the output values.

3. How can I find the critical points of the function?

To find the critical points, set the derivative of the function equal to zero and solve for x. In this case, the derivative is (x^3 + 3x + 12)' = 3x^2 + 3, which gives critical points at x = -1 and x = 1.

4. How can I determine if the function has any local extrema?

To determine the local extrema, evaluate the second derivative of the function at the critical points. If the second derivative is positive, the function has a local minimum at that point. If the second derivative is negative, the function has a local maximum. In this case, the second derivative is 6x, which gives a local minimum at x = -1 and a local maximum at x = 1.

5. Can this function be integrated using elementary functions?

Yes, this function can be integrated using elementary functions. The resulting integral is (x^2 + 6ln(x) - 6ln(x+2)) + C.

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