# How do you find the integral of (x^4+x-4) / (x^2+2)?

First, use polynomial long division to get $\frac{{x}^{4} + x - 4}{{x}^{2} + 2} = {x}^{2} - 2 + \frac{x}{{x}^{2} + 2}$. Next, integrate this expression, using a substitution $u = {x}^{2} + 2 , \mathrm{du} = 2 x \setminus \mathrm{dx}$ to get:
$\setminus \int \frac{{x}^{4} + x - 4}{{x}^{2} + 2} \mathrm{dx} = \setminus \int \left({x}^{2} - 2 + \frac{x}{{x}^{2} + 2}\right) \mathrm{dx}$
$= \setminus \frac{1}{3} {x}^{3} - 2 x + \setminus \frac{1}{2} \ln \left({x}^{2} + 2\right) + C$
Note that absolute value signs are not needed in the argument of the logarithm function since ${x}^{2} + 2 > 0$ for all real $x$.