How would I integrate: $\displaystyle \int_0^2 \! \dfrac{2x}{x^2+4} \, dx.$
How would I integrate: $\displaystyle \int_0^2 \! \dfrac{2x}{x^2+4} \, dx.$
Hint: the integrand is of the form $\displaystyle \frac{u'(x)}{u(x)}$...