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

Nov 25, 2016

#### Explanation:

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

$\int \frac{{x}^{2} - x + 9}{{x}^{2} + 9} ^ 2 \mathrm{dx} = \int \frac{{x}^{2} + 9}{{x}^{2} + 9} ^ 2 \mathrm{dx} - \int \frac{x}{{x}^{2} + 9} ^ 2 \mathrm{dx}$

$\int \frac{{x}^{2} - x + 9}{{x}^{2} + 9} ^ 2 \mathrm{dx} = \int \frac{1}{{x}^{2} + 9} \mathrm{dx} - \frac{1}{2} \int \frac{2 x}{{x}^{2} + 9} ^ 2 \mathrm{dx}$

The first integral is our old friend the inverse tangent and the second integral is all set up for a "u" substitution. Let $u = {x}^{2} + 9 , \text{then } \mathrm{du} = 2 x \mathrm{dx}$

$\frac{{x}^{2} - x + 9}{{x}^{2} + 9} ^ 2 = \frac{A}{{x}^{2} + 9} + \frac{B x + C}{{x}^{2} + 9} ^ 2$

${x}^{2} - x + 9 = A \left({x}^{2} + 9\right) + B x + C$

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) ]

$A = 1 , B = - 1 \mathmr{and} C = 0$

$\int \frac{{x}^{2} - x + 9}{{x}^{2} + 9} ^ 2 \mathrm{dx} = \int \frac{1}{{x}^{2} + 9} \mathrm{dx} - \int \frac{x}{{x}^{2} + 9} ^ 2 \mathrm{dx}$

$\int \frac{{x}^{2} - x + 9}{{x}^{2} + 9} ^ 2 \mathrm{dx} = \int \frac{1}{{x}^{2} + 9} \mathrm{dx} - \frac{1}{2} \int \frac{2 x}{{x}^{2} + 9} ^ 2 \mathrm{dx}$

Let $u = {x}^{2} + 9 , \text{then } \mathrm{du} = 2 x \mathrm{dx}$

$\int \frac{{x}^{2} - x + 9}{{x}^{2} + 9} ^ 2 \mathrm{dx} = \int \frac{1}{{x}^{2} + 9} \mathrm{dx} - \frac{1}{2} \int \frac{1}{u} ^ 2 \mathrm{du}$

Integrate

$\int \frac{{x}^{2} - x + 9}{{x}^{2} + 9} ^ 2 \mathrm{dx} = \frac{1}{3} {\tan}^{-} 1 \left(\frac{x}{3}\right) + \frac{1}{2} \frac{1}{u} + C$

Reverse the substitution:

$\int \frac{{x}^{2} - x + 9}{{x}^{2} + 9} ^ 2 \mathrm{dx} = \frac{1}{3} {\tan}^{-} 1 \left(\frac{x}{3}\right) + \frac{1}{2 \left({x}^{2} + 9\right)} + C$