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A couple pretty easy integration problems im stuck on

  1. Apr 21, 2004 #1
    integral of 10/((x-1)(x^2+9)) dx

    integral of x^3/((x+1)^3) dx

    both these are under the partial fraction section, so using those methods would be helpful...

    thanks!
     
  2. jcsd
  3. Apr 21, 2004 #2
    The function 10/((x-1)(x^2+9) can be broken up into the partial fractions

    1/(x-1) - (x+1)/(x^2+9)

    The first one is simple (natural log of abs(denominator)). The second part is a bit more involved. I used trig substitution and worked through it. Use x=3*tan(theta).

    When all is said there will most likely be more than one correct possible answer but the one I found is

    ln(abs(x-1))-ln(Sqrt(x^2+9))-(1/3)*arctan(x/3) plus a constant.



    For the second one the first thing I did was to do long division. This yielded

    1- (3x^2+3x+1)/(x+1)^3

    Next I worked out the partial fractions with the remainder that is left over from long division.

    A/(x+1) + B/(x+1)^2 + c/(x+1)^3 = (3x^2+3x+1)/(x+1)^3

    Some algebra (I like to equate the coefficients for this) yielded

    A=3, B=-3 and C=1

    Then do the integration on each section and you should get

    x - 3*ln(abs(x+1)) - 3/(x+1) + 1/(2*(x+1)^2) plus a constant of course.
    And if you take the derivative of this you will get back you integrand, which means you are correct.
     
    Last edited: Apr 21, 2004
  4. Apr 21, 2004 #3
    I got... for the 1st one

    ln|x+1|-0.5ln|x^2+9|-1/3*tan^-1(x/3)+C

    2nd one...

    x+1-3ln|x+1|-3/(x+1)-1/(2*(x+1)^2)+C

    thats all I think...
     
  5. Apr 21, 2004 #4
    I did substitution... and did u=x+1 and went from there for the 2nd one...
     
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