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

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Answer 1 · 2016-02-08T03:23:15

#int frac{1}{x(x+1)^2} dx = ln|x| - ln|x+1| + 1/(x+1) + "Constant"#

As stated by the question, we must first express #frac{1}{x(x+1)^2}# in partial fractions, before we perform the integration.

#frac{1}{x(x+1)^2} -= A/x + B/(x+1) + C/(x+1)^2#,

where #A#, #B# and #C# are constants to be determined, such that the equality holds true for all possible values of #x#.

Which means,

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

Simplifying gives

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

We compare the coefficients (to zero) to get 3 simultaneous linear equations,

#A+B=0#
#2A+B+C=0#
#A=1#

The system of equations is solved easily to yield

#A=1#
#B=-1#
#C=-1#

Therefore,

#frac{1}{x(x+1)^2} -= 1/x - 1/(x+1) - 1/(x+1)^2#.

Now, to integrate.

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

#= ln|x| - ln|x+1| + 1/(x+1) + "Constant"#