Intergration of Rational Functions (Multiple Qs)

[irok]

Evaluate the Integral:

\int \frac {2x+1}{(x^{2}+9)^{2}}

My attempt:

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

= (Ax+B)(x^{2} + 9)^{2} + (Cx+D)(x^{2} + 9)

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

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

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

2x + 1 = (Ax+B)(x^{2}+9) + Cx + D

Still not sure what I'm doing wrong.

 

[n_bourbaki]

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.

 

[irok]

Evaluate the integral:
\int \frac {-2x^{2} - 9x - 50}{x^{3} + 8x^{2} + 30x + 36}

x^{3} + 8x^{2} + 30x + 36 = (x+2)(x^{2}+6x+18)

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

I'm stuck at this point. I tried finding roots for x^{2}+6x+18 but can't. tried dividing x^{2}+6x+18 with numerator but there was a remainder of 3x-14.

 

[rocomath]

Use Partial Fractions.

 

[HallsofIvy]

Evaluate the Integral:

\int \frac {2x+1}{(x^{2}+9)^{2}}

My attempt:

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

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

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

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
2x+ 1= Ax^{5} + Bx^{4} Dx^{3} + (18A + E)x^{2} + (81A+9D+18B)x + 9E + 81B
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:
\frac {2x+1}{(x^{2}+9)^{2}} = \frac {Ax+B}{x^{2}+9} + \frac {Cx+D}{(x^{2} + 9)^{2}}

2x + 1 = (Ax+B)(x^{2}+9) + Cx + D

Still not sure what I'm doing wrong.

 

[HallsofIvy]

Evaluate the integral:
\int \frac {-2x^{2} - 9x - 50}{x^{3} + 8x^{2} + 30x + 36}

x^{3} + 8x^{2} + 30x + 36 = (x+2)(x^{2}+6x+18)

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

I'm stuck at this point. I tried finding roots for x^{2}+6x+18 but can't. tried dividing x^{2}+6x+18 with numerator but there was a remainder of 3x-14.

Again, use partial fractions.
\frac{-2x^2- 9x+ 50}{(x+2)(x^2+ 6x+ 18)}= \frac{A}{x+2}+ {Bx+ C}{x^2+ 6x+ 18}
so
-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 /> x^2+ 6x+ 18 cannot be factored but you can complete the square: x^2+ 6x+ 18= x^2+ 6x+ 9+ 9= (x+3)^2+ 9<br /> To do that integral, let u= x+ 3 so the denominator is u²+ 9.