# okintergration of 4x-3/(2x-1)^3

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### 1 Answer

You need to evaluate the indefinite integral such that:

`int (4x - 3)/((2x - 1)^3)dx`

You need to use the property of linearity of integral, hence, you need to split the integral in two simpler integrals, such that:

`int (4x - 3)/((2x - 1)^3)dx = int (4x)/((2x - 1)^3)dx - int 3/((2x - 1)^3)dx`

You need to use partial fraction decomposition to evaluate `int (4x)/((2x - 1)^3)dx` , such that:

`(4x)/((2x - 1)^3) = A/(2x - 1) + B/((2x - 1)^2) + C/((2x - 1)^3)`

`4x = A(2x - 1)^2 + B(2x - 1) + C`

`4x = 4Ax^2 - 4Ax + A + 2Bx - B + C`

`4x = 4Ax^2 + x(-4A + 2B) + A - B + C`

Equating the coefficients of like powers, yields:

`4A = 0 => A = 0`

`-4A + 2B = 4 => 2B = 4 => B = 2`

`A - B + C = 0 => 0 - 2 + C = 0 => C = 2`

`(4x)/((2x - 1)^3) = 0/(2x - 1) + 2/((2x - 1)^2) + 2/((2x - 1)^3)`

Integrating both sides, yields:

`int (4x)/((2x - 1)^3)dx = int 2/((2x - 1)^2) dx + int 2/((2x - 1)^3) dx`

`int (4x - 3)/((2x - 1)^3)dx = int 2/((2x - 1)^2) dx + int 2/((2x - 1)^3) dx - int 3/((2x - 1)^3)dx`

`int (4x - 3)/((2x - 1)^3)dx = int 2/((2x - 1)^2) dx - int 1/((2x - 1)^3)dx`

You should come up with the substitution such that:

`2x - 1 = u => 2dx = du => dx = (du)/2`

`int 2/((2x - 1)^2) dx = int (du)/(u^2) = -1/u + c`

`int 2/((2x - 1)^2) dx = -1/(2x - 1) + c`

`int 1/((2x - 1)^3)dx = -1/(2(2x - 1)^2) + c`

`int (4x - 3)/((2x - 1)^3)dx = -1/(2x - 1) - 1/(2(2x - 1)^2) + c`

**Hence, evaluating the given indefinite integral yields **`int (4x - 3)/((2x - 1)^3)dx = -1/(2x - 1) - 1/(2(2x - 1)^2) + c.`