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