Solving the Integral 2x^3-2x^2+1/x^2-x-2

  • Thread starter kennis2
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  • #1
kennis2
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i can't resolve this integral :confused:
2x^3-2x^2+1/x^2-x-2?
 
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  • #2
Is that

[tex]\frac{2x^3-2x^2+1}{x^2-x-2}[/tex]

?

If so, use polynomial longdivision and then partial fractions to transform the expression into something which can be integrated more easily.
 
  • #3
yeah, can you so kind to do some procedures.. because i did with partial fractions
but the result is not correct =(
thx much!
 
  • #4
Partial fractions don't work because the denominator is less than the numerator, use Muzza's idea and use long division.
 
  • #5
Long division is part of the partial fractions algorithm!
 
  • #6
[tex]\int \frac{2x^3-2x^2+1}{x^2-x-2}[/tex]
[tex]\frac{2x^3-2x^2+1}{x^2-x-2}=2x+\frac{2x+1}{2x^3-2x^2+1}[/tex]
[tex]\int 2xdx + \int \frac{2x+1}{2x^3-2x^2+1}dx[/tex]
now, use partial fractions on the second half. it shouldn't be too hard!
 
  • #7
The denominator is [itex](x-2)(x+1) [/itex],so it shouldn't be 2 difficult to get the simple fractions...



Daniel.
 
  • #8
p53ud0 dr34m5 said:
[tex]\frac{2x^3-2x^2+1}{x^2-x-2}=2x+\frac{2x+1}{2x^3-2x^2+1}[/tex]

Did you mean

[tex]\frac{2x^3-2x^2+1}{x^2-x-2}=2x+\frac{4x+1}{x^2-x-2}[/tex]?
 
  • #9
Yes,of course,yours is correct.He couldn't have changed the denominator.I mean he did,but it was wrong.

Daniel.
 
  • #10
[itex]\frac{2x^3-2x^2+1}{x^2-x-2}[/itex]=[itex]2x+\frac{4x+1}{x^2-x-2}[/itex]
This is correct.

Partial Fraction Decompisition for [itex]\frac{4x+1}{x^2-x-2}[/itex] is:


[itex]\frac{4x+1}{(x+1)(x-2)}[/itex] = [itex]\frac{1}{x+1} + \frac{3}{x-2}[/itex]

New Integral is:
[itex]\int (2x + \frac{1}{x+1} + \frac{3}{x-2})dx[/itex]

so,

[itex]\int(\frac{2x^3-2x^2+1}{x^2-x-2})dx = x^2 + ln|x+1| + 3*ln|x-2| + C[/itex]
 
  • #11
x^2 + 3 Log[2 - x] + Log[1 + x]
 

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