# Question ce174

Mar 22, 2017

You perform partial fraction decomposition and then integrate each of the terms.

#### Explanation:

Here is the basic partial fraction equation:

$\frac{3 {x}^{4} + {x}^{3} + 20 {x}^{2} + 3 x + 31}{\left(x + 1\right) {\left({x}^{2} + 4\right)}^{2}} = \frac{A}{x + 1} + \frac{B x + C}{{x}^{2} + 4} + \frac{D x + E}{{\left({x}^{2} + 4\right)}^{2}}$

Multiply both sides by $\left(x + 1\right) {\left({x}^{2} + 4\right)}^{2}$

$3 {x}^{4} + {x}^{3} + 20 {x}^{2} + 3 x + 31 = A {\left({x}^{2} + 4\right)}^{2} + \left(B x + C\right) \left(x + 1\right) \left({x}^{2} + 4\right) + \left(D x + E\right) \left(x + 1\right)$

We need 5 equations.

Let $x = - 1$:

$3 {\left(- 1\right)}^{4} + {\left(- 1\right)}^{3} + 20 {\left(- 1\right)}^{2} + 3 \left(- 1\right) + 31 = A {\left({\left(- 1\right)}^{2} + 4\right)}^{2} + \left(B \left(- 1\right) + C\right) \left(- 1 + 1\right) \left({\left(- 1\right)}^{2} + 4\right) + \left(D \left(- 1\right) + E\right) \left(- 1 + 1\right)$

$3 {\left(- 1\right)}^{4} + {\left(- 1\right)}^{3} + 20 {\left(- 1\right)}^{2} + 3 \left(- 1\right) + 31 = A {\left({\left(- 1\right)}^{2} + 4\right)}^{2} + \left(B x + C\right) \left(- 1 + 1\right) \left({x}^{2} + 4\right) + \left(D x + E\right) \left(- 1 + 1\right)$

$50 = 25 A + 0 B + 0 C + 0 D + 0 E$

$A = 2$

The first line in the augmented matrix is:

$\left[\left(1 , 0 , 0 , 0 , 0 , | , 2\right)\right]$

Let $x = 0$:

31 = $3 {\left(0\right)}^{4} + {\left(0\right)}^{3} + 20 {\left(0\right)}^{2} + 3 \left(0\right) + 31 = A {\left({\left(0\right)}^{2} + 4\right)}^{2} + \left(B \left(0\right) + C\right) \left(0 + 1\right) \left({\left(0\right)}^{2} + 4\right) + \left(D \left(0\right) + E\right) \left(0 + 1\right)$

$31 = 16 A + 0 B + 4 C + 0 D + E$

The next line in the augmented matrix is:

[(1,0,0,0,0,|,2), (16,0,4,0,1,|,31) ]

Let x = 1:

31 = $3 {\left(1\right)}^{4} + {\left(1\right)}^{3} + 20 {\left(1\right)}^{2} + 3 \left(1\right) + 31 = A {\left({\left(1\right)}^{2} + 4\right)}^{2} + \left(B \left(1\right) + C\right) \left(1 + 1\right) \left({\left(1\right)}^{2} + 4\right) + \left(D \left(1\right) + E\right) \left(1 + 1\right)$

$58 = 25 A + 10 B + 10 C + 2 D + 2 E$

[(1,0,0,0,0,|,2), (16,0,4,0,1,|,31), (25,10,10,2,2,|,58) ]

This is enough to show you how to do it.
You solve the matrix and obtain values for all 5 variables
You find that A = 2, B = 1, C = 0 , D=0, and E=-1

Here is the results broken into 3 integrals:

$\int \frac{3 {x}^{4} + {x}^{3} + 20 {x}^{2} + 3 x + 31}{\left(x + 1\right) {\left({x}^{2} + 4\right)}^{2}} \mathrm{dx} = 2 \int \frac{1}{x + 1} \mathrm{dx} + \int \frac{x}{{x}^{2} + 4} \mathrm{dx} - \int \frac{1}{{x}^{2} + 4} ^ 2 \mathrm{dx}$

The first integral is the natural logarithm:

$\int \frac{3 {x}^{4} + {x}^{3} + 20 {x}^{2} + 3 x + 31}{\left(x + 1\right) {\left({x}^{2} + 4\right)}^{2}} \mathrm{dx} = 2 \ln \left(x + 1\right) + \int \frac{x}{{x}^{2} + 4} \mathrm{dx} - \int \frac{1}{{x}^{2} + 4} ^ 2 \mathrm{dx}$

For the second integral let $u = {x}^{2} + 4$, then $\mathrm{du} = 2 x \mathrm{dx}$ and it, too, becomes a natural logarithm:

$\int \frac{3 {x}^{4} + {x}^{3} + 20 {x}^{2} + 3 x + 31}{\left(x + 1\right) {\left({x}^{2} + 4\right)}^{2}} \mathrm{dx} = 2 \ln \left(x + 1\right) + \frac{1}{2} \ln \left({x}^{2} + 4\right) - \int \frac{1}{{x}^{2} + 4} ^ 2 \mathrm{dx}$

For the third integral, let $x = 2 \tan \left(u\right)$, and $\mathrm{dx} = 2 {\sec}^{2} \left(u\right) \mathrm{du}$ and it becomes the following:

$\int \frac{3 {x}^{4} + {x}^{3} + 20 {x}^{2} + 3 x + 31}{\left(x + 1\right) {\left({x}^{2} + 4\right)}^{2}} \mathrm{dx} = 2 \ln \left(x + 1\right) + \frac{1}{2} \ln \left({x}^{2} + 4\right) - \frac{1}{16} \left(\frac{2 x}{{x}^{2} + 4} + {\tan}^{-} 1 \left(\frac{x}{2}\right)\right) + C$