How do you integrate #int (5x^3+2x^2-12x-8)/(x^4-8x^2+16) dx# using partial fractions?

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Answer 1 · 2016-12-04T08:14:28

The answer is #=1/(x+2)+2ln(∣x+2∣)-1/(x-2)+3ln(∣x-2∣)+C#

We use,

#(a-b)^2=a^2-2ab+b^2#

and #a^2-b^2=(a+b)(a-b)#

to factorise the denominator

#x^4-8x^2+16=(x^2-4)^2#

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

So,

#(5x^3+2x^2-12x-8)/(x^4-8x^2+16)=(5x^3+2x^2-12x-8)/((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)#

So,

#5x^3+2x^2-12x-8=A(x-2)^2+B(x+2)(x-2)^2+C(x+2)^2+D(x-2)(x+2)^2#

Let #x=2#, #=>#, #16=16C#, #=>#, #C=1#

Let #x=-2#, #=>#, #-16=16A#, #=>#, #A=-1#

Coefficients of #x^3#, #5=B+D#

And #-8=4A+8B+4C-8D#

#-8=-4+8B+4-8D#

#-8=8B-8D#

#B-D=-1#

So, #B=2# and #D=3#

#int((5x^3+2x^2-12x-8)dx)/(x^4-8x^2+16)#

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

#=1/(x+2)+2ln(∣x+2∣)-1/(x-2)+3ln(∣x-2∣)+C#