Intergration of Rational Functions (Multiple Qs)

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Homework Help Overview

The discussion revolves around the integration of rational functions, specifically focusing on the use of partial fractions to evaluate integrals. Participants are attempting to solve integrals involving polynomials in the numerator and denominator.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to decompose rational functions into partial fractions but express confusion regarding their setups and calculations. Some are questioning the correctness of their initial assumptions and the steps taken in their attempts.

Discussion Status

There are multiple attempts to evaluate integrals, with some participants providing feedback on the methods used. Guidance has been offered regarding the use of partial fractions and the need to equate coefficients to solve for unknowns. However, there is no explicit consensus on the correct approach, and participants are exploring different interpretations of the problems.

Contextual Notes

Some participants mention difficulties in finding roots for quadratic expressions and express uncertainty about the correctness of their algebraic manipulations. There is also a note about the inability to factor certain polynomials, which may affect the integration process.

irok
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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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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.
 
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.
 
Use Partial Fractions.
 
irok said:
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.
 
irok said:
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]<br /> choose 3 values for x to get 3 equations for A, B, and C. Or multiply out the right side and equate corresponding coefficients.<br /> [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]<br /> To do that integral, let u= x+ 3 so the denominator is u<sup>2</sup>+ 9.[/tex]
 

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