How do you use partial fractions to find the integral #int (6x^2+1)/(x^2(x-1)^3)dx#?

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Answer 1 · 2017-10-12T17:00:50

#-3lnx+1/x-3ln(x-1)+2/(x-1)-7/2*(x-1)^(-2)+C#

I decomposed integrand into basic fractions,

#(6x^2+1)/[x^2*(x-1)^2]#

=#-3/x-1/x^2+3/(x-1)-2/(x-1)^2+7/(x-1)^3#

Hence,

#int ((6x^2+1)dx)/[x^2*(x-1)^3]#

=#-int (3dx)/x#-#int dx/x^2#-#int (3dx)/(x-1)#-#int (2dx)/(x-1)^2#+#int (7dx)/(x-1)^3#

=#-3lnx+1/x-3ln(x-1)+2/(x-1)-7/2*(x-1)^(-2)+C#