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Partial fractions

  1. Mar 15, 2010 #1
    [tex]\int \frac{2x+1}{4x^2+12x-7}dx[/tex]
    [tex]\frac{1}{4} \int \frac{2x+1}{x^2+3x-\frac{7}{4}}dx[/tex]
    [tex]\frac{1}{4} \int \frac{2x+1}{(x+\frac{3}{2})^2-4}dx[/tex]
    [tex]u=x+\frac{3}{2}[/tex]
    [tex]\frac{1}{2} \int \frac{u-1}{u^2-4}du[/tex]
    [tex]\frac{1}{2} \int \frac{u}{u^2-4}du -\frac{1}{2}\int \frac{du}{u^2-4}[/tex]
    [tex]\frac{1}{4} ln|u^2-4|-\frac{1}{2}\int \frac{A}{u+2} +\frac{B}{u-2} du[/tex]
    [tex]-\frac{1}{2}=A(u-2)+B(u+2)[/tex]
    [tex]A=\frac{1}{8}[/tex]
    [tex]B=-\frac{1}{8}[/tex]
    [tex]\frac{1}{4} ln|u^2-4|+\frac{1}{8}ln|\frac{u+2}{u-2}|+C[/tex]
    [tex]\frac{1}{4} ln|x^2+3x-\frac{7}{4}|-\frac{1}{8} ln|\frac{x+\frac{7}{2}}{x-\frac{1}{2}}|+C[/tex]
     
  2. jcsd
  3. Mar 15, 2010 #2

    gabbagabbahey

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    Homework Helper
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    What you've got is correct, but it can still be simplified further; [itex]\ln|u^2-4|=\ln|u-2|+\ln|u-2|[/itex] and [itex]\ln\left|\frac{u+2}{u-2}\right|=\ln|u-2|-\ln|u-2|[/itex].

    So,

    [tex]\frac{1}{4}\ln|u^2-4|+\frac{1}{8}ln\left|\frac{u+2}{u-2}\right|=\frac{3}{8}\ln|u+2|+\frac{1}{8}\ln|u-2|[/tex]
     
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