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Integration by parts/partial fractions

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

    gabbagabbahey

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    Not quite,

    [tex]\int\frac{2u^2}{u^2+\frac{7}{4}}du\neq \ln|u^2+\frac{7}{4}|[/tex]
     
  4. Mar 21, 2010 #3
    eh,
    [tex]
    -\int \frac{2u^2}{u^2+\frac{7}{4}}du=\int 2-\frac{7}{2(u^2+\frac{7}{4})}du
    [/tex]
    [tex]
    -2u+\frac{7}{\sqrt{7}}tan^{-1}\frac{2u}{\sqrt{7}}+C
    [/tex]
    [tex]
    (x-\frac{1}{2})ln(x^2-x+2)-2x+1+\frac{7}{\sqrt{7}}tan^-1\frac{2x-1}{\sqrt{7}}
    [/tex]
     
  5. Mar 21, 2010 #4

    gabbagabbahey

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    That looks more or less correct.:approve: You are, of course, missing a constant of integration here, and since 1 is also a constant you might as well absorb it into the integration constant and write it as

    [tex]\int \ln\left(x^2-x+2\right)dx=\left(x-\frac{1}{2}\right)\ln\left(x^2-x+2\right)-2x+\sqrt{7}\tan^{-1}\left(\frac{2x-1}{\sqrt{7}}\right)+C[/tex]
     
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