Here is the basic partial fraction equation:
(3x^4+x^3+20x^2+3x+31)/((x+1)(x^2+4)^2) = A/(x+1) + (Bx+C)/(x^2+4)+(Dx+E)/((x^2+4)^2)3x4+x3+20x2+3x+31(x+1)(x2+4)2=Ax+1+Bx+Cx2+4+Dx+E(x2+4)2
Multiply both sides by (x+1)(x^2+4)^2(x+1)(x2+4)2
3x^4+x^3+20x^2+3x+31 = A(x^2+4)^2 + (Bx+C)(x+1)(x^2+4)+(Dx+E)(x+1)3x4+x3+20x2+3x+31=A(x2+4)2+(Bx+C)(x+1)(x2+4)+(Dx+E)(x+1)
We need 5 equations.
Let x = -1x=−1:
3(-1)^4+(-1)^3+20(-1)^2+3(-1)+31 = A((-1)^2+4)^2 + (B(-1)+C)(-1+1)((-1)^2+4)+(D(-1)+E)(-1+1)3(−1)4+(−1)3+20(−1)2+3(−1)+31=A((−1)2+4)2+(B(−1)+C)(−1+1)((−1)2+4)+(D(−1)+E)(−1+1)
3(-1)^4+(-1)^3+20(-1)^2+3(-1)+31 = A((-1)^2+4)^2 + (Bx+C)(-1+1)(x^2+4)+(Dx+E)(-1+1)3(−1)4+(−1)3+20(−1)2+3(−1)+31=A((−1)2+4)2+(Bx+C)(−1+1)(x2+4)+(Dx+E)(−1+1)
50 = 25A+0B+0C+0D+0E50=25A+0B+0C+0D+0E
A = 2A=2
The first line in the augmented matrix is:
[(1,0,0,0,0,|,2)]
Let x = 0:
31 = 3(0)^4+(0)^3+20(0)^2+3(0)+31 = A((0)^2+4)^2 + (B(0)+C)(0+1)((0)^2+4)+(D(0)+E)(0+1)
31 = 16A + 0B+ 4C + 0D+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(1)^4+(1)^3+20(1)^2+3(1)+31 = A((1)^2+4)^2 + (B(1)+C)(1+1)((1)^2+4)+(D(1)+E)(1+1)
58 = 25A + 10B+ 10C + 2D+2E
[(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(3x^4+x^3+20x^2+3x+31)/((x+1)(x^2+4)^2)dx = 2int1/(x+1)dx+intx/(x^2+4)dx-int1/(x^2+4)^2dx
The first integral is the natural logarithm:
int(3x^4+x^3+20x^2+3x+31)/((x+1)(x^2+4)^2)dx = 2ln(x+1)+intx/(x^2+4)dx-int1/(x^2+4)^2dx
For the second integral let u = x^2+4, then du = 2xdx and it, too, becomes a natural logarithm:
int(3x^4+x^3+20x^2+3x+31)/((x+1)(x^2+4)^2)dx = 2ln(x+1)+1/2ln(x^2+4)-int1/(x^2+4)^2dx
For the third integral, let x = 2tan(u), and dx = 2sec^2(u)du and it becomes the following:
int(3x^4+x^3+20x^2+3x+31)/((x+1)(x^2+4)^2)dx = 2ln(x+1)+1/2ln(x^2+4)-1/16((2x)/(x^2+4)+ tan^-1(x/2)) +C
