By reducing the improper fraction and using a substitution to reduce it to standard form evaluate 4x^3-x^2+16x/x^2+4 dx
the function f = (4x^3-x^2+16x)/(x^2+4 ) = 4x - 1 + 4/(x^2 + 4)
I = int f dx = int (4x - 1) dx + int 4/(x^2 + 4) dx
I = 2x^2 - x + I2, where I2 = int 4/(x^2 + 4) dx
Using the substitution x = 2tan t with dx = 2sec^2 t dt
I2 = int 4/(x^2 + 4) dx = int 4/{4tan^2 t + 4} dx = int 4/{4(tan^2 t + 1)} * (2sec^2 t dt)
I2 = int 1/(sec^2 t) * (2sec^2 t dt) = int 2 dt = 2t
I2 = 2t = 2arctan(x/2)
Final integral is: I = 2x^2 - x + 2arctan(x/2)