How do you use partial fractions to find the integral #int (x^3)/(x^2-4)^2dx#?

Question

Charles Arocha · Accepted Answer

The answer is #=1/2ln(|x^2-4|)-2/(x^2-4)+C#

Perform the decomposition into partial fractions

#x^3/(x^2-4)^2=(x^3)/((x-2)^2(x+2)^2)#

#=A/(x-2)^2+B/(x-2)+C/(x+2)^2+D/(x+2)#

#=(A(x+2)^2+B(x-2)(x+2)^2+C(x-2)^2+D(x+2)(x-2)^2)/((x-2)^2(x+2)^2)#

The denominators are the same, compare the numerators

#x^3=A(x+2)^2+B(x-2)(x+2)^2+C(x-2)^2+D(x+2)(x-2)^2#

Let #x=2#, #=>#, #8=16A#, #=>#, #A=1/2#

Let #x=-2#, #=>#, #-8=16C#, #=>#, #C=-1/2#

Coefficients of #x^3#

#1=B+D#

Coefficients of #x^2#

#0=A+2B+C-2D#

#B-D=0#

#B=B=1/2#

#x^3/(x^2-4)^2=(1/2)/(x-2)^2+(1/2)/(x-2)+(-1/2)/(x+2)^2+(1/2)/(x+2)#

Therefore, the integral is

#int(x^3dx)/(x^2-4)^2=int(1/2dx)/(x-2)^2+int(1/2dx)/(x-2)+int(-1/2dx)/(x+2)^2+int(1/2dx)/(x+2)#

#=-1/2*1/(x-2)+1/2ln(|x-2|)+1/2*1/(x+2)+1/2ln(|x+2|)+C#

#=1/2ln(|x^2-4|)-2/(x^2-4)+C#

Charles Arocha · Answer

# 1/2ln|(x^2-4)|-2/(x^2-4)+C#.

As a Second Method, let us solve the Problem without applying the Method of Partial Fraction. Consider the subst. #x^2=y," so that, "2xdx=dy#. # :. I=intx^3/(x^2-4)^2dx=1/2int(x^2*2x)/(x^2-4)^2dx#, #=1/2inty/(y-4)^2dy#, #=1/2int{(y-4)+4}/(y-4)^2dy#, #=1/2int{(y-4)/(y-4)^2+4/(y-4)^2}dy#, #=1/2{int1/(y-4)dy+4int1/(y-4)^2dy}#, #=1/2{ln|(y-4)|+4*(y-4)^(-2+1)/(-2+1)}#. #=1/2ln|(y-4)|-2/(y-4)#. Returning to #x^2=y#, we get, # I=1/2ln|(x^2-4)|-2/(x^2-4)+C#, as Respected Narad T. has readily derived! Enjoy Maths.!

William Abdalla · Answer

undefined To integrate $ \frac{x^3}{(x^2-4)^2} $ using partial fractions, first factor the denominator:

$ x^2 - 4 = (x - 2)(x + 2) $

Then, express the fraction as the sum of partial fractions:

$ \frac{x^3}{(x^2-4)^2} = \frac{A}{x-2} + \frac{B}{x+2} + \frac{Cx + D}{(x^2-4)} + \frac{Ex + F}{(x^2-4)^2} $

Clear the denominators and solve for $ A $, $ B $, $ C $, $ D $, $ E $, and $ F $.

Next, integrate each partial fraction separately.

Finally, integrate the resulting expressions and add the constants of integration.