How do you evaluate the integral #int 1/(x(x-1)^3)#?

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Answer 1 · 2018-03-14T21:24:01

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

Write:

#1/(x(x-1)^3) = A/x+B/(x-1)+C/(x-1)^2+D/(x-1)^3#

Multiplying both sides by #x(x-1)^3# we get:

#1 = A(x-1)^3+Bx(x-1)^2+Cx(x-1)+Dx#

Putting #x=0# we find #A=-1#

Looking at the coefficient of #x^3#, we find #B=-A = 1#

Putting #x=1# we find #D=1#

Looking at the coefficient of #x#, we find:

#0 = 3A+B-C+D#

Hence #C=-1#

So

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

and:

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

#color(white)(int 1/(x(x-1)^3) dx) = -ln abs(x)+ln abs(x-1)+1/(x-1)-1/(2(x-1)^2)+C#