How do you integrate #int ( (dx) / ( x(x+1)^2 ) )# using partial fractions?

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Answer 1 · 2017-11-02T16:40:26

The partial fraction equation is:

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

Multiply both sides by #x(x+1)^2#

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

Eliminate B and C by Letting #x = 0#:

#1 = A(0+1)^2#

#A = 1#

#1 = (x+1)^2 + Bx(x+1)+Cx#

Eliminate B by letting #x = -1#

#1 = C(-1)#

#C = -1#

#1 = (x+1)^2 + Bx(x+1)-x#

Let #x = 1#:

#1 = (1+1)^2 + B(1)(1+1)-1#

#2-4 = 2B#

#B = -1#

The partial fraction expansion is:

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

Check:

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

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

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

This checks.

The original integrand is equal to the partial fractions:

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

Separate into 3 integrals:

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

These integrals are well known:

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