Intergration of Rational Functions (Multiple Qs)

In summary: Then use the quadratic formula with the correct coefficients: \int_{-2}^+\infty \frac{du}{du^2}=\frac{1}{2}u^2+\frac{1}{4}u+\frac{1}{8}u^4+\ldots+\frac{1}{16}u^8+\frac{1}{32}u^16+\ldots=\frac{1}{64}u^64[/itex]
  • #1
irok
13
0
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.
 
Last edited:
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  • #2
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
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
Use Partial Fractions.
 
  • #5
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.
 
  • #6
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]
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.
 

What is the definition of integration of rational functions?

Integration of rational functions is the process of finding the antiderivative of a rational function. It involves finding a function whose derivative is equal to the given rational function.

Can all rational functions be integrated?

No, not all rational functions can be integrated. Some rational functions are considered to be "improper" and cannot be integrated using the standard methods. In these cases, more advanced techniques such as partial fraction decomposition may be used.

What are some common techniques used to integrate rational functions?

Some common techniques used to integrate rational functions include substitution, integration by parts, and partial fraction decomposition. Different techniques may be more suitable for different types of rational functions.

How does integration of rational functions relate to finding areas under curves?

Integration of rational functions is closely related to finding areas under curves. By finding the antiderivative of a rational function, we can determine the area under the curve of that function within a given interval.

What are some real-world applications of integration of rational functions?

Integration of rational functions has many real-world applications, especially in fields such as physics, engineering, and economics. It can be used to solve problems involving rates of change, optimization, and finding areas under curves in real-world scenarios.

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