Find `int(2x^3 + 11x^2 +28x +33)/(x^2 - x +6)dx` using the partial fraction method.

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You need to divide `2x^3 + 11x^2 + 28x + 33` by `x^2 - x + 6` using reminder theorem, such that:

`(2x^3 + 11x^2 + 28x + 33)/(x^2 - x + 6) = 2x + 13 + (29x - 45)/(x^2 - x + 6)`

Integrating both sides yields:

`int (2x^3 + 11x^2 + 28x + 33)/(x^2 - x + 6) dx = int 2x dx + 13 int dx + int (29x - 45)/(x^2 - x + 6) dx`

`int (2x^3 + 11x^2 + 28x + 33)/(x^2 - x + 6) dx = 2x^2/2 + 13x + 29 int (x - 45/29)/(x^2 - x + 6) dx`

You need to solve the integral `int (x - 45/29)/(x^2 - x + 6) dx` , hence, you need to subtract 1 and add 1, such that:

`(1/2)int (2x - 1 + 1 - 90/29)/(x^2 - x + 6) dx`

You should come up with the substitution, such that:

`x^2 - x + 6 = t => (2x - 1)dx = dt`

`(1/2)int (2x - 1 + 1 - 90/29)/(x^2 - x + 6) dx = (1/2)int (2x - 1)/(x^2 - x + 6) dx + (1/2)int (1 - 90/29)/(x^2 - x + 6) dx`

`(1/2)int (2x - 1 + 1 - 90/29)/(x^2 - x + 6) dx = (1/2) int (dt)/t - 61/58 int 1/(x^2 - x + 6) dx`

You should use the following formula of integration, such that:

`int 1/(x^2 + a^2) dx = (1/a) tan^(-1)(x/a)` , hence, you should complete the square to denominator such that:

`(x^2 - x + 1/4) - 1/4 + 6 = (x - 1/2)^2 + 23/4`

`int 1/(x^2 - x + 6) dx = int 1/((x - 1/2)^2 + 23/4) dx`

You need to come up with the substitution, such that:

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

`int 1/((u)^2 + 23/4) du = 2/sqrt23 tan^(-1)(2u/sqrt23) + c`

Substituting back `x - 1/2` for u yields:

`int 1/((x - 1/2)^2 + 23/4) dx = 2/sqrt23 tan^(-1)((2x - 1)/sqrt23) + c`

`(1/2)int (2x - 1 + 1 - 90/29)/(x^2 - x + 6) dx = (1/2)ln(x^2 - x + 6) - 61/(29sqrt 23)tan^(-1)((2x - 1)/sqrt23) + c`

`int (2x^3 + 11x^2 + 28x + 33)/(x^2 - x + 6) dx = 2x^2/2 + 13x + 29/2ln(x^2 - x + 6) -61/(sqrt 23)tan^(-1)((2x - 1)/sqrt23) + c`

Hence, evaluating the given integral yields` int (2x^3 + 11x^2 + 28x + 33)/(x^2 - x + 6) dx = 2x^2/2 + 13x + 29/2ln(x^2 - x + 6) -61/(sqrt 23)tan^(-1)((2x - 1)/sqrt23) + c.`

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