Question #99cca

2 Answers
Jan 7, 2018

#1/2x^2-2x-1/2ln|x+1|+1/6ln|x-1|+16/3ln|x+2|+C#

Explanation:

#intx^4/(x^3+2x^2-x-2)dx#

#=int(x^4+2x^3-x^2-2x)/(x^3+2x^2-x-2)+(-2x^3+x^2+2x)/(x^3+2x^2-x-2)dx# (split fraction so the first fraction will be divisible by the denominator. this reduces the power of the numerator of the resulting fraction from 4 to 3)

#=intx+(-2x^3-4x^2+2x+4)/(x^3+2x^2-x-2)+(5x^2-4)/(x^3+2x^2-x-2)dx# (split fraction again)

#=1/2x^2+int-2+(5x^2-4)/((x^2-1)(x+2))dx# (power rule integration and factorize denominator)

#=1/2x^2-2x+int(5x^2-4)/((x+1)(x-1)(x+2))dx#

now use partial fraction decomposition for the fraction:

#(5x^2-4)/((x+1)(x-1)(x+2))=A/(x+1)+B/(x-1)+C/(x+2)# (A, B, and C are constants)

#5x^2-4=A(x-1)(x+2)+B(x+1)(x+2)+C(x+1)(x-1)#

if x=-1:

#5(-1)^2-4=A(-1-1)(-1+2)+B(-1+1)(-1+2)+C(-1+1)(-1-1)#
#1=A(-2)(1)+B(0)(1)+C(0)(-2)#
#1=A(-2)#
#-1/2=A#

if x=1:

#5(1)^2-4=A(1-1)(1+2)+B(1+1)(1+2)+C(1+1)(1-1)#
#1=A(0)(3)+B(2)(3)+C(2)(0)#
#1=B(6)#
#1/6=B#

if x=-2:

#5(-2)^2-4=A(-2-1)(-2+2)+B(-2+1)(-2+2)+C(-2+1)(-2-1)#
#16=A(-3)(0)+B(-1)(0)+C(-1)(-3)#
#16=C(3)#
#16/3=C#

substitute:
#=1/2x^2-2x+int(-1/2)/(x+1)+(1/6)/(x-1)+(16/3)/(x+2)dx#
#=1/2x^2-2x-1/2ln|x+1|+1/6ln|x-1|+16/3ln|x+2|+C# (integration to natural log)

Jan 7, 2018

The answer is #=x^2/2-2x+16/3ln(|x+2|)-1/2ln(|x+1|)+1/6ln(|x-1|)+C#

Explanation:

The degree of the numerator is greater than the degree of the denominator, perform a polynomial long division.

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

Factorise the denominator,

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

#x^4/(x^3+2x^2-x-2)=(x-2)+((5x^2-4))/((x+2)(x+1)(x-1))#

Perform a decompostion into partial fractions

#((5x^2-4))/((x+2)(x+1)(x-1))=A/(x+2)+B/(x+1)+C/(x-1)#

#=(A(x+1)(x-1)+B(x+2)(x-1)+C(x+2)(x+1))/((x+2)(x+1)(x-1))#

The denominator is the same, compare the numerator

#5x^2-4=A(x+1)(x-1)+B(x+2)(x-1)+C(x+2)(x+1)#

Let #x=-2#, #=>#, #16=3A#, #=>#, #A=16/3#

Let #x=-1#, #=>#, #1=-2B#, #=>#, #B=-1/2#

Let #x=1#, #=>#, #1=6C#, #=>#, #C=1/6#

Therefore,

#x^4/(x^3+2x^2-x-2)=(x-2)+(16/3)/(x+2)-(1/2)/(x+1)+(1/6)/(x-1)#

Perform the integration

#int(x^4dx)/(x^3+2x^2-x-2)=int(x-2)dx+int(16/3dx)/(x+2)-int(1/2dx)/(x+1)+int(1/6dx)/(x-1)#

#=x^2/2-2x+16/3ln(|x+2|)-1/2ln(|x+1|)+1/6ln(|x-1|)+C#