2du/1+3u How do you integrate this?
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Move the 2 outside the integral then use substitution where v=1+3u. You should get (2/3)ln(abs(1+3u)).
Originally Posted by kinana18 2du/1+3u How do you integrate this? $\displaystyle \int \frac{2}{1+3u} du = \frac{2}{3} \int \frac{3}{1+3u} du = \frac{2}{3} \ln|1 + 3u| + C $
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