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Integration by parts/partial fractions

  • Thread starter nameVoid
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  • #1
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[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]
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
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[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]
Not quite,

[tex]\int\frac{2u^2}{u^2+\frac{7}{4}}du\neq \ln|u^2+\frac{7}{4}|[/tex]
 
  • #3
241
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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]
 
  • #4
gabbagabbahey
Homework Helper
Gold Member
5,002
6
[tex]
(x-\frac{1}{2})ln(x^2-x+2)-2x+1+\frac{7}{\sqrt{7}}tan^-1\frac{2x-1}{\sqrt{7}}
[/tex]
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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