How do you use partial fractions to find the integral #int (x^2-x+9)/(x^2+9)^2dx#?

1 Answer
Nov 25, 2016

Please see the explanation.

Explanation:

I would not use partial fractions. Here is what I would do:

#int(x^2 - x + 9)/(x^2 + 9)^2 dx = int(x^2 + 9)/(x^2 + 9)^2dx - int(x)/(x^2 + 9)^2dx#

#int(x^2 - x + 9)/(x^2 + 9)^2 dx = int1/(x^2 + 9)dx - 1/2int(2x)/(x^2 + 9)^2dx#

The first integral is our old friend the inverse tangent and the second integral is all set up for a "u" substitution. Let #u = x^2 + 9, "then "du = 2xdx#

But since you asked:

#(x^2 - x + 9)/(x^2 + 9)^2 = A/(x^2 + 9) + (Bx + C)/(x^2 + 9)^2#

#x^2 - x + 9 = A(x^2 + 9) + Bx + C#

Let x = 0:

#[ (9, 0, 1,|,9) ]#

Let x = 1:

#[ (9, 0, 1,|,9), (10,1,1,|,9) ]#

Let x = -1:

#[ (9, 0, 1,|,9), (10,1,1,|,9), (10,-1,1,|,11) ]#

Subtract row 3 from row 2:

#[ (9, 0, 1,|,9), (0,2,0,|,-2), (10,-1,1,|,11) ]#

Divide row 2 by 2:

#[ (9, 0, 1,|,9), (0,1,0,|,-1), (10,-1,1,|,11) ]#

Subtract row 3 from row 1:

#[ (-1, 1, 0,|,-2), (0,1,0,|,-1), (10,-1,1,|,11) ]#

Subtract row 2 from row 1:

#[ (-1, 0, 0,|,-1), (0,1,0,|,-1), (10,-1,1,|,11) ]#

Multiply row 1 by 10 and add to row 3:

#[ (-1, 0, 0,|,-1), (0,1,0,|,-1), (0,-1,1,|,1) ]#

Multiply row 1 by -1

#[ (1, 0, 0,|,1), (0,1,0,|,-1), (0,-1,1,|,1) ]#

Add row 2 to row 3:

#[ (1, 0, 0,|,1), (0,1,0,|,-1), (0,0,1,|,0) ]#

#A = 1, B = -1 and C = 0#

#int(x^2 - x + 9)/(x^2 + 9)^2 dx = int1/(x^2 + 9)dx - intx/(x^2 + 9)^2dx#

#int(x^2 - x + 9)/(x^2 + 9)^2 dx = int1/(x^2 + 9)dx - 1/2int(2x)/(x^2 + 9)^2dx#

Let #u = x^2 + 9, "then "du = 2xdx#

#int(x^2 - x + 9)/(x^2 + 9)^2 dx = int1/(x^2 + 9)dx - 1/2int1/u^2du#

Integrate

#int(x^2 - x + 9)/(x^2 + 9)^2 dx = 1/3tan^-1(x/3) + 1/2(1)/u + C#

Reverse the substitution:

#int(x^2 - x + 9)/(x^2 + 9)^2 dx = 1/3tan^-1(x/3) + (1)/(2(x^2 + 9)) + C#