How do you integrate #int 2/(x^3-x^2)# using partial fractions?

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Answer 1 · 2016-12-16T22:08:55

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

#x^3 - x^2# can be factored as #x^2(x - 1)#.

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

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

#Ax^2 + Bx - Ax - B + Cx^2 = 2#

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

We can now write a system of equations.

#{(A + C = 0), (B - A = 0), (-B = 2):}#

Solving, we obtain #B = -2#, #A = -2#, #C = 2#.

Therefore, the partial fraction decomposition is:

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

We now decompose #(-2x - 2)/x^2# into partial fractions.

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

#A + Bx = -2x - 2 -> A = -2 and B = -2#

Therefore, the complete partial fraction decomposition of #2/(x^3 - x^2)# is #-2/x^2 - 2/x + 2/(x - 1)#.

This can be integrated as follows:

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

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

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

#=2ln|x- 1| - 2ln|x| + 2/x + C#

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

Hopefully this helps!