Question

Evaluate the integral x^3-3x^2+4x-10/x^2+4 dx, I got to the point where i need to take the integral of (x-3) + integral of (2/x^2+4). But i'm not sure if that's right?

Answer 1

I think this is just about what you want to do.\[\int\limits (x^3-3x^2+4x-10)/(x^2+4) dx=\int\limits (x-3+2/(x^2+4))dx\]\[=x^2/2-3x+\tan ^{-1}(x/2)+C\]

Answer 2

Those inverse trig functions are a little tricky.