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Integrating Rational Functions

  1. Jul 20, 2012 #1
    1. ∫[itex]\frac{dx}{x3 + 2x}[/itex]

    We're suppose to evaluate the integral.

    Use Partial Fraction Decomposition:

    [itex]\frac{1}{x3 + 2x}[/itex] = [itex]\frac{A}{x}[/itex] + [itex]\frac{Bx + C}{x2 + 2}[/itex]

    1 = A(x2 + 2) + (Bx + C)(x)
    1 = Ax2 + 2A + Bx2 + Cx
    1 = x2( A + B) + Cx + 2A

    Solving for A gives [itex]\frac{1}{2}[/itex]
    Solving for B gives -[itex]\frac{1}{2}[/itex]
    Solving for C gives 0

    ∫[itex]\frac{dx}{x(x2 + 2}[/itex] = [itex]\frac{1}{2}[/itex]∫[itex]\frac{dx}{x}[/itex] - [itex]\frac{1}{2}[/itex]∫[itex]\frac{dx}{x2 + 2}[/itex]

    When we evaluate this, I get:

    [itex]\frac{1}{2}[/itex]ln x - [itex]\frac{1}{2}[/itex]tan-1[itex]\frac{x}{\sqrt{2}}[/itex]

    Or should it be:

    [itex]\frac{1}{2}[/itex]ln x - [itex]\frac{1}{2}[/itex]ln (x2 + 2)
     
  2. jcsd
  3. Jul 20, 2012 #2

    Mark44

    Staff: Mentor

    B was the coefficient of x in the numerator. What happened to x?

    Also, don't mix tags inside of [itex] expressions. It is causing what you wrote to not render correctly.

     
  4. Jul 20, 2012 #3


    I fixed it:

    And I got

    \frac{1}{2}ln x - \frac{1}{4}ln(x2 + 2)

    Is this correct?
     
  5. Jul 20, 2012 #4

    Mark44

    Staff: Mentor

    Why don't you check for yourself? If your answer is correct, it should be true that
    d/dx[1/2 * ln(x) - (1/4) * ln(x2 + 2)] = 1/(x3 + 2x), the original integrand.
     
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