Evaluate the integral integrate of ((x^3-4x^2+3x+1)/(x^2+4))dx

Since the
degree of numerator is larger than degree of denominator, you need to use reminder theorem such
that:

`A(x) = B(x)C(x) + R(x)`

Notice that C(x) represents
the quotient and R(x) represents the reminder.

`x^3-4x^2+3x+1 = (x^2+4)(ax+b)
+ cx + d`

`x^3-4x^2+3x+1 = ax^3 + bx^2 + 4ax + 4b + cx + d`

`x^3-4x^2+3x+1 = ax^3 + bx^2 + x(4a+c) + 4b + d`

Equating the
coefficients of like powers yields:

`a = 1`

`b =
-4`

`4a + c = 3 => c = -1`

`4b + d = 1 => -16 + d =
1 => d = 17`

`x^3-4x^2+3x+1 = (x^2+4)(x-4) - x + 17`

Dividing both sides by `x^2 + 4`  yields:

`(x^3-4x^2+3x+1)/(x^2+4) =
x - 4 + (- x + 17)/(x^2+4)`

Integrating both sides yields:

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

`int (x^3-4x^2+3x+1)/(x^2+4) dx = x^2/2 - 4x - int x/(x^2+4)dx
+17 int 1/(x^2+4) dx`

You should use the following substitution to solve `int
x/(x^2+4) dx ` such that:

`x^2 + 4 = t => 2xdx = dt => xdx =
(dt)/2`

`int x/(x^2+4) dx= (1/2)int (dt)/t = (1/2)ln|t|+c`

`int (x^3-4x^2+3x+1)/(x^2+4) dx = x^2/2 - 4x - (1/2)ln(x^2+4) + 17/2*arctan (x/2) +
c`

Hence, evaluating the given integral yields `int
(x^3-4x^2+3x+1)/(x^2+4) dx = x^2/2 - 4x - (1/2)ln(x^2+4) + 17/2*arctan (x/2) + c.`