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:

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

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

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