How do you integrate #int (2x+1) /( (x-2)(x^2+4))# using partial fractions?

1 Answer
Sep 30, 2017

#((5-ln|x-2|)-(3*tan^-1(x/2)))/8+(5ln|x^2+4|)/16+c#

Explanation:

  1. First step in this problem is understanding that the use of partial fractions is valid since the numerator is a lesser degree of power than the denominator.

  2. Apply Partial Fractions:
    *Note that the denominator terms are Linear (left) and Quadratic (on the right), all this means is that they are terms that cannot be factorized further.
    Linear form = #A/(x-2)#
    Quadratic form = #(Bx+C)/(x^2+4)#

Add these two terms together and set it equal to the Numerator.
#2x+1=A/(x-2)+(Bx+C)/(x^2+4)#

Simplify by multiplying A by the denomenator of the other fraction, and the same for B:
#2x+1=A(x^2+4)+Bx+C(x-2)#

now distribute:
#2x+1=Ax^2+4A+Bx^2-2Bx+Cx-2C#

*Note: The biggest thing students confuse about partial fractions is this next step, in the left hand side you should think of it as #0x^2+2x+1# to understand that there are Zero #x^2#'s on the left.

#color(red)(0x^2)+color(blue)(2x)+color(green)(1)=color(red)(Ax^2)+color(green)(4A)+color(red)(Bx^2)-color(blue)(2Bx)+color(blue)(Cx)-color(green)(2C)#

Collect like terms:
#color(red)(x^2) rArr# #0=A+B#
#color(blue)x rArr# #2=-2B+C#
#color(green)"Integers"# #rArr# #1=4A-2C#

Solve by any method, here we will use substitution
Eq1: #0=A+B rArr [A=-B] hArr [B=-A]#
Eq2: #2=-2B+C rArr [2+2B=C]#
Eq3: #1=4A-2C rArr 1=2(2A-C) rArr [1/2=2A-C]#

Substitute in Eq1 and Eq2 into Eq3 to get:
#1/2 = 2(-B)-(2+2B)#
Simplify:
#1/2=-4B-2#

Solve for B:
#B=-5/8#
Use B to solve for A
#A=5/8#
Use B to solve for C
#2+2(-5/8)=C#
#C=3/4#

Plug in A, B, and C values into original fraction:
#A/(x-2)+(Bx+C)/(x^2+4)#

#(5/8)/(x-2)+(-5/8x+3/4)/(x^2+4)#

Simplify into:
#5/(8(x-2))+(-5x+6)/(8(x^2+4))#

New integral is now represented by:
#int5/(8(x-2))dx+int(-5x+6)/(8(x^2+4))dx#

Remove constants to prepare for integration:
#5*int1/(8(x-2))dx+1/8*int(-5x+6)/((x^2+4))dx#

Break apart second integral (on the right) into two separate integrals (Mind the 1/8 multiplier) and remove the constants in the numerators

#5/8*int1/((x-2))dx+1/8*(-5*intx/(x^2+4)dx+6*int1/(x^2+4)dx)#

First integral:
#5/8int1/(x-2)dx#
#u=x-2#
Result is: #(5ln|x-2|)/8#

Second Integral:
#5*intx/(x^2+4)dx#
#u=x^2+4#
Result Is: #(-5ln|x^2+4|)/2#

Third Integral:
#6*int1/(x^2+4)dx#
Apply inverse trig formula by identifying denominator as Arctan:
#int1/(x^2+a^2)dx = 1/a * tan^-1(x/a)#
Where #[a^2=4] rArr [a=2]#
Result is: #3tan^-1(x/2)#

Plug these back into full equation and do not forget the #1/8# multiplier for the second and third integrals.

#(5*ln|x-2|)/8-1/8((-5ln|x^2+4|)/2+3tan^-1(x/2))#

Distribute the #1/8# multiplier

#(5*ln|x-2|)/8+(5ln|x-2|)/16-(3tan^-1(x/2))/8+c#

Simplify:
#((5-ln|x-2|)-(3*tan^-1(x/2)))/8+(5ln|x^2+4|)/16+c#

You can simplify further if you like, however this is an acceptable answer.