Question #ce174

1 Answer
Mar 22, 2017

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

Explanation:

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)#

Multiply both sides by #(x+1)(x^2+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)#

We need 5 equations.

Let #x = -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)#

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

#A = 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#