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Math Help - Integration problem

  1. #1
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    Integration problem

    \int\frac{5x^2}{x^3-1}dx
    Hi I'm trying to integrate the above function. Not sure where to begin, I cant seem to break it up into partial fractions.
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  2. #2
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    Rewrite it as \displaystyle \frac{5}{3}\int{\frac{3x^2}{x^3 - 1}\,dx}, then make the substitution \displaystyle u = x^3 - 1.
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  3. #3
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    Quote Originally Posted by flashylightsmeow View Post
    \int\frac{5x^2}{x^3-1}dx
    Hi I'm trying to integrate the above function. Not sure where to begin, I cant seem to break it up into partial fractions.
    You don't need partial fractions at all here: observe that \displaystyle{5x^2=\frac{5}{3}\frac{d(x^3-1)}{dx} , so you get an immediate logarithmic integral:

    \displaystyle{\int\frac{5x^2}{x^3-1}\,dx=\frac{5}{3}\int\frac{3x^2}{x^3-1}\,dx , and the integral's of the form \displaystyle{\int\frac{f'}{f}\,dx=\ln|f|+K ,

    but if you insist in partial fractions then:


    \displaystyle{x^3-1=(x-1)(x^2+x+1)\Longrightarrow \frac{5x^2}{x^3-1}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}\iff}

    \displaystyle{\iff 5x^2=A(x^2+x+1)+(Bx+C)(x-1)}

    Now choose nice values of x (say, 0, 1 ,-1) and solve resp. to find the coefficientes A,B,C.

    At the end the integral turns out to be pretty easy: it equals \displaystyle{\frac{5}{3}\ln|x^3-1|+K\,,\,\,K=\mbox{ a constant}}

    Tonio
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  4. #4
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    Yes, partial fractions is the hard way- but it can be done. x^3- 1= (x- 1)(x^2+ x+ 1) so that
    \frac{5x^2}{x^3- 1}= \frac{A}{x-1}+ \frac{Bx+ C}{x^2+ x+ 1}

    5x^2= A(x^2+ x+ 1)+ (Bx+ C)(x- 1)

    Taking x= 1, 5= 3A so A= 5/3. Taking x= 0, 0= (5/3)(1)+ C(-1) and C= -5/3. Taking x= -1, 5= 5/3+ (-5/3- B)(-2)= 2B+ 15/3= 2B+ 5 so B= 0.

    Your integral is
    \frac{5}{3}\int \frac{dx}{x-1}- \frac{5}{3}\int \frac{dx}{x^2+ x+ 1}

    The first is easy to integrate. To integrate the second, complete the square: x^2+ x+ 1= x^2+ x+ 1/4+ 3/4= (x+1/2)^2+ 3/4 so the integral is
    [tex]-\frac{5}{3}\int \frac{dx}{(x+1/2)^2+ 3/4}[/itex]
    and the substitution u= x+ 1/2 makes that
    [tex]-\frac{5}{3}\int \frac{du}{u^2+ 3/4}= -\frac{20}{9}\int\frac{du}{(4u^2}/3+ 1}[/itex]
    and the final substitution v= 2u/\sqrt{3} makes it
    -\frac{10\sqrt{3}}{9}\int \frac{dv}{v^2+ 1}
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    Yes, partial fractions is the hard way- but it can be done. x^3- 1= (x- 1)(x^2+ x+ 1) so that
    \frac{5x^2}{x^3- 1}= \frac{A}{x-1}+ \frac{Bx+ C}{x^2+ x+ 1}

    5x^2= A(x^2+ x+ 1)+ (Bx+ C)(x- 1)

    Taking x= 1, 5= 3A so A= 5/3. Taking x= 0, 0= (5/3)(1)+ C(-1) and C= -5/3.


    Caution: it should be \displaystyle{C=\frac{5}{3}

    Tonio


    Taking x= -1, 5= 5/3+ (-5/3- B)(-2)= 2B+ 15/3= 2B+ 5 so B= 0.

    Your integral is
    \frac{5}{3}\int \frac{dx}{x-1}- \frac{5}{3}\int \frac{dx}{x^2+ x+ 1}

    The first is easy to integrate. To integrate the second, complete the square: x^2+ x+ 1= x^2+ x+ 1/4+ 3/4= (x+1/2)^2+ 3/4 so the integral is
    [tex]-\frac{5}{3}\int \frac{dx}{(x+1/2)^2+ 3/4}[/itex]
    and the substitution u= x+ 1/2 makes that
    [tex]-\frac{5}{3}\int \frac{du}{u^2+ 3/4}= -\frac{20}{9}\int\frac{du}{(4u^2}/3+ 1}[/itex]
    and the final substitution v= 2u/\sqrt{3} makes it
    -\frac{10\sqrt{3}}{9}\int \frac{dv}{v^2+ 1}
    .
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