How do you use partial fractions to find the integral #int (x+2)/(x^2-4x)dx#?

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Answer 1 · 2016-12-28T16:10:11

First, factor the denominator.

#x^2 -4x = x(x - 4)#

#A/x + B/(x - 4) = (x + 2)/((x)(x - 4))#

#A(x -4) + B(x) = x + 2#

#Ax+ Bx - 4A = x + 2#

#(A + B)x - 4A = x + 2#

We can now write a systems of equations.

#{(A + B = 1),(-4A = 2):}#

Solving, we get #A = -1/2# and #B = 3/2#.

#:.# The partial fraction decomposition is #3/(2(x - 4)) - 1/(2x)#.

This can be integrated using the rule #int1/udu = ln|u| + C#.

#=3/2ln|x - 4| - 1/2ln|x| + C#

Hopefully this helps!