Intergration of Rational Functions (Multiple Qs)

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  • #1
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Evaluate the Integral:

[tex] \int \frac {2x+1}{(x^{2}+9)^{2}}[/tex]

My attempt:

[tex] \frac {2x+1}{(x^{2}+9)^{2}} = \frac {Ax+B}{x^{2}+9} + \frac {Cx+D}{(x^{2} + 9)^{2}}[/tex]

= [tex] (Ax+B)(x^{2} + 9)^{2} + (Cx+D)(x^{2} + 9) [/tex]

= [tex] Ax^{5} + Bx^{4} Dx^{3} + (18A + E)x^{2} + (81A+9D+18B)x + 9E + 81B[/tex]

I'm not sure what I'm doing wrong here since I can't find value of A B C or D.

2nd attempt:
[tex] \frac {2x+1}{(x^{2}+9)^{2}} = \frac {Ax+B}{x^{2}+9} + \frac {Cx+D}{(x^{2} + 9)^{2}}[/tex]

[tex] 2x + 1 = (Ax+B)(x^{2}+9) + Cx + D [/tex]

Still not sure what I'm doing wrong.
 
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Answers and Replies

  • #2
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Well, all those things you've written as if there were equalities are clearly not equal.


In the the first line after "My attempt:" where is the denominator on the LHS? Ditto on the second line for the RHS. Why have you multiplied Ax+B and Cx+D those powers of x^2 + 9?

I think you need to start all over again with this attempt at the use of partial fractions.
 
  • #3
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Evaluate the integral:
[tex] \int \frac {-2x^{2} - 9x - 50}{x^{3} + 8x^{2} + 30x + 36}[/tex]

[tex] x^{3} + 8x^{2} + 30x + 36 = (x+2)(x^{2}+6x+18) [/tex]

[tex] \int \frac {-2x^{2} - 9x - 50}{(x+2)(x^{2}+6x+18)}[/tex]

I'm stuck at this point. I tried finding roots for [tex] x^{2}+6x+18 [/tex] but can't. tried dividing [tex] x^{2}+6x+18 [/tex] with numerator but there was a remainder of 3x-14.
 
  • #4
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Use Partial Fractions.
 
  • #5
HallsofIvy
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Homework Helper
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Evaluate the Integral:

[tex] \int \frac {2x+1}{(x^{2}+9)^{2}}[/tex]

My attempt:

[tex] \frac {2x+1}{(x^{2}+9)^{2}} = \frac {Ax+B}{x^{2}+9} + \frac {Cx+D}{(x^{2} + 9)^{2}}[/tex]

= [tex] (Ax+B)(x^{2} + 9)^{2} + (Cx+D)(x^{2} + 9) [/tex]
You mean
[tex]2x+ 1= (Ax+B)(x^2+ 9)^2+ (Cx+D)(x^2+ 9)[/tex]

= [tex] Ax^{5} + Bx^{4} Dx^{3} + (18A + E)x^{2} + (81A+9D+18B)x + 9E + 81B[/tex]

I'm not sure what I'm doing wrong here since I can't find value of A B C or D.
Again, that should be
[tex]2x+ 1= Ax^{5} + Bx^{4} Dx^{3} + (18A + E)x^{2} + (81A+9D+18B)x + 9E + 81B[/tex]
Since that is true for all x, taking 4 convenient values for x will give you 4 equations for A, B, C, D.

2nd attempt:
[tex] \frac {2x+1}{(x^{2}+9)^{2}} = \frac {Ax+B}{x^{2}+9} + \frac {Cx+D}{(x^{2} + 9)^{2}}[/tex]

[tex] 2x + 1 = (Ax+B)(x^{2}+9) + Cx + D [/tex]

Still not sure what I'm doing wrong.
 
  • #6
HallsofIvy
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Evaluate the integral:
[tex] \int \frac {-2x^{2} - 9x - 50}{x^{3} + 8x^{2} + 30x + 36}[/tex]

[tex] x^{3} + 8x^{2} + 30x + 36 = (x+2)(x^{2}+6x+18) [/tex]

[tex] \int \frac {-2x^{2} - 9x - 50}{(x+2)(x^{2}+6x+18)}[/tex]

I'm stuck at this point. I tried finding roots for [tex] x^{2}+6x+18 [/tex] but can't. tried dividing [tex] x^{2}+6x+18 [/tex] with numerator but there was a remainder of 3x-14.
Again, use partial fractions.
[tex]\frac{-2x^2- 9x+ 50}{(x+2)(x^2+ 6x+ 18)}= \frac{A}{x+2}+ {Bx+ C}{x^2+ 6x+ 18}[/tex]
so
[tex]-2x^2- 9x+ 50= A(x^2+ 6x+ 18)+ (Bx+ C)(x+ 2)[/itex]
choose 3 values for x to get 3 equations for A, B, and C. Or multiply out the right side and equate corresponding coefficients.
[itex]x^2+ 6x+ 18[/itex] cannot be factored but you can complete the square: [itex]x^2+ 6x+ 18= x^2+ 6x+ 9+ 9= (x+3)^2+ 9[/itex]
To do that integral, let u= x+ 3 so the denominator is u2+ 9.
 

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