Simplify the integrand as
[tex]\dfrac{2x^3 - x^2 - 2x + 4}{1 + x^2} = \dfrac{(2x^3 + 2x) - (x^2 + 1) - 4x + 5}{x^2 + 1} \\\\ = \dfrac{2x(x^2 + 1) - (x^2 + 1) - 4x + 5}{x^2 + 1} \\\\ = 2x - 1 - \dfrac{4x - 5}{x^2 + 1}[/tex]
(in other words, compute the quotient and remainder)
We can further split up and prepare the remainder term for integration by rewriting it as
[tex]\dfrac{4x - 5}{x^2 + 1} = 2\times\dfrac{2x}{x^2 + 1} - \dfrac5{x^2 + 1}[/tex]
Now we integrate:
[tex]\displaystyle \frac{2x^3 - x^2 - 2x + 4}{1 + x^2} \, dx = \int \left(2x - 1 - 2\times\frac{2x}{x^2+1} + \frac5{x^2+1}\right) \, dx[/tex]
[tex]\displaystyle = x^2 - x - 2 \int \frac{2x}{x^2+1} \, dx + 5 \int \frac{dx}{x^2+1}[/tex]
In the first remaining integral, substitute y = x² + 1 and dy = 2x dx. In the last integral, recall that d/dx [arctan(x)] = 1/(x² + 1).
[tex]\displaystyle = x^2 - x - 2 \int \frac{dy}y + 5 \int \frac{dx}{x^2+1}[/tex]
[tex]\displaystyle = x^2 - x - 2 \ln|y| + 5 \arctan(x) + C[/tex]
[tex]\displaystyle = \boxed{x^2 - x - 2 \ln(x^2+1) + 5 \arctan(x) + C}[/tex]
