How do you use partial fractions to find the integral #int (5x^2-12x-12)/(x^3-4x)dx#?

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Answer 1 · 2017-11-04T23:57:23

#int ((5x^2-12x-12)*dx)/(x^3-4x)#=#4Ln(x+2)+3Lnx-2Ln(x-2)+C#

#(5x^2-12x-12)/(x^3-4x)#

=#(5x^2-12x-12)/[x*(x+2)*(x-2)]#

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

After expanding denominator,

#A*x*(x-2)+B*(x+2)*(x-2)+C*x*(x+2)=5x^2-12x-12#

Set #x=-2#, #8A=32#, so #A=4#

Set #x=0#, #-4B=-12#, so #B=3#

Set #x=2#, #-8C=-16#, so #C=-2#

Thus,

#int ((5x^2-12x-12)*dx)/(x^3-4x)#

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

=#4Ln(x+2)+3Lnx-2Ln(x-2)+C#