$\frac{1}{x^2+1}$
I thought it would be $\frac{1}{2x}*ln(x^2+1)$, because it equals $\frac{1}{x^2+1}$. But I got $\frac{1}{8}ln5$=.20 and when I checked the definite integral by calculator it was 1.325. What should the antiderivative be?