How do you integrate # (3x^2 + 12x - 20 )/(x^4 - 8x^2 + 16)# using partial fractions?

Question

3091 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-07T06:56:46

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

#(3x^2+12x-20)/(x^4-8x^2+16)=(3x^2+12x-20)/((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,

#3x^2+12x-20=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#, #=>#, #-32=16A#, #=>#, #A=-2#

Coefficients of #x^3#, #0=B+D#

And #-20=4A+8B+4C-8D#

#-20=-8+8B+4-8D#

#-16=8B-8D#

#B-D=-2#

So, #B=-1# and #D=1#

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

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

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