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.`

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

Rewrite the integrand using long division:

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

`=int xdx-4int dx+17int(dx)/(x^2+4)-intx/(x^2+4)dx`

We will add the constant of integration at the end:

`int xdx=1/2x^2`

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`-4intdx=-4x`

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`int x/(x^2+4)` Let `u=x^2+4,du=2xdx` so we get

`int x/(x^2+4)=1/2int(du)/u=1/2ln|u|=1/2ln|x^2+4|`

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`int(dx)/(x^2+4)` is in the form `int (du)/(a^2+u^2)=1/(a) arctan(u/a)` where `a=2,u=x` so `int(dx)/(x^2+4)=1/(2) arctan(x/2)`

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All together we have:

`int(x^3-4x^2+3x+1)/(x^2+4) dx=1/2x^2 -4x-1/2ln|x^2+4|+17[1/(2) arctan(x/2)]+C`

You can factor out the i/2 to get:

`=1/2[x^2-8x-ln|x^2+4|+17 arctan(x/2)]+C`