# Intergration of Rational Functions (Multiple Qs)

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.

Last edited:

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

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

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

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.

Use Partial Fractions.

HallsofIvy
Homework Helper
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
Homework Helper
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
[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.
$x^2+ 6x+ 18$ cannot be factored but you can complete the square: $x^2+ 6x+ 18= x^2+ 6x+ 9+ 9= (x+3)^2+ 9$
To do that integral, let u= x+ 3 so the denominator is u2+ 9.