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Calc II Partial Fractions

  1. Sep 13, 2011 #1
    [itex]\int\frac{7x^2+4x+5}{x^3+x} dx[/itex]

    I have to solve this using partial fractions and all I could think of was breaking it up into:


    also I tried

    The reasoning behind this one was to split up original fraction into [itex]\frac{7x^2+4x+5}{(x^3+0)+x} dx[/itex]

    Are either of these two even right? and if so I have no clue how to proceed.

    Forgive me for not showing more work, but I have been staring at this for 30 minutes and can think of nothing. Thank you.
  2. jcsd
  3. Sep 13, 2011 #2


    Staff: Mentor

    For irreducible quadratic factors in the denominator (like x2 + 1), the numerator should be Ax + B. IOW, your decomposition should be:
    [tex]\frac{A}{x} + \frac{Bx + C}{x^2+1}[/tex]
  4. Sep 13, 2011 #3
    Try breaking it up into

  5. Sep 13, 2011 #4
    Alright so then if I multiplied through by x would i have:

    [tex]\frac{7x^2+4x+5}{x^2+1}=A + \frac{x(Bx + C)}{x^2+1}[/tex]
  6. Sep 13, 2011 #5


    Staff: Mentor

    You need to solve for A, B, and C, and these values do not depend on x. Just do the algebra first.
    [tex]\frac{7x^2+4x+5}{x^3+x} =\frac{A}{x} + \frac{Bx + C}{x^2+1}[/tex]

    You're solving for A, B, and C so that this equation is an identity - true for all values of x other than those that make the denominators zero.

    The first thing to do is to multiply both sides by x(x2 + 1)
  7. Sep 13, 2011 #6
    When I solve for A, B and C - I get A=5, B=7, and C=4.

    For the final answer i get:

    [tex]5ln(x)+4arctan(x)+\frac{7}{2}ln(x^2+1)[/tex] when the real answer is [tex]5ln(x)+4arctan(x)+ln(x^2+1)[/tex] Thanks for helping me get almost to the end but can you help me figure out why my answer is different? I solved for B by dividing both sides by [tex]x^2[/tex]
  8. Sep 13, 2011 #7
    Not to be a nuisance but I still need help.
  9. Sep 13, 2011 #8
    Yea this stuff just gets a little getting used to. What do you need help with still? Are you unclear of the decomposing fractions?
  10. Sep 13, 2011 #9
    Thanks, this is what I don't understand:

  11. Sep 13, 2011 #10
    I think you screwd up a bit on the algebra.
    A + B=7
  12. Sep 13, 2011 #11
    Alright, is the equation

    [tex]7x^2+4x+5=A(x^2+1)+Bx^2+Cx[/tex] correct for when I multiply both sides by [tex]x(x^2+1)[/tex] I'm kinda lost here for how I would solve for B. Wouldn't I divide by x^2, then set x=0, and that would give B=7?
  13. Sep 13, 2011 #12

    if 7x^2 + 4x +5 = Ax^2 + A + Bx^2 +Cx

    then how do you find the coefficients of A, B and C? (hint: the coefficients are directly associated with the powers of exponents).
  14. Sep 13, 2011 #13
    I get C=4 and A=5 easy. I can not for the life of be understand why B does not equal 7. I have no clue what your hint means. Why can't you divide both sides by x^2 and get B=7 if you set x=0??
  15. Sep 13, 2011 #14


    Staff: Mentor

    First off, if you set x = 0, you can't then come around and divide by x2.
    The equation you are solving is
    7x2 + 4x + 5 = (A + B)x2 + Cx + A
    This has to be identically true, so the coefficients of x2 and x and the constants have to be equal.
  16. Sep 13, 2011 #15
    Thank you for you help, I think I have a better understanding of solving when the denominator is irreducible but my real problem was that the denominator in this particular problem was x^3-x not x^3+x..... thank you all for your help, however.
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