$\displaystyle \frac{1}{x^2+1}$

bounds 0 to 4.

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