How do you find #int (x-1) / ((2x+1) (4x+4)) dx# using partial fractions?

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Answer 1 · 2016-10-22T01:43:50

Please see the explanation.

Please notice that the integral can be multiplied by #1/4# and the integrand becomes:

#(x -1)/((2x + 1)(x + 1)) = A/(2x + 1) + B/(x + 1)#

Multiply by the left denominator:

#x -1 = A(x + 1) + B(2x + 1)#

Let #x = -1#

#-2 = -B#

#B = 2#

Let (x = -1/2)

#-3/2 = A(1/2)#

#A = -3#

Check:

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

#(4x +2 - 3x - 3)/((2x + 1)(x + 1)) = #

#(x - 1)/((2x + 1)(x + 1)) # This checks:

#int(x -1)/((2x + 1)(4x + 4))dx = 1/2int1/(x + 1)dx - 3/4int 1/(2x + 1)dx#

#int(x -1)/((2x + 1)(4x + 4))dx = 1/2ln|x + 1| - 3/8ln|2x + 1| + C#