$\displaystyle \int \frac{2x}{e^x}dx$ I tried: u=2x dv=$\displaystyle e^{-x}dx$ but I got a funky answer. Is this the correct substitution? Still learning which to use when...lol
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Originally Posted by saiyanmx89 $\displaystyle \int \frac{2x}{e^x}dx$ I tried: u=2x dv=$\displaystyle e^{-x}dx$ but I got a funky answer. Is this the correct substitution? Still learning which to use when...lol Yes.
Originally Posted by saiyanmx89 $\displaystyle \int \frac{2x}{e^x}dx$ I tried: u=2x dv=$\displaystyle e^{-x}dx$ but I got a funky answer. Is this the correct substitution? Still learning which to use when...lol $\displaystyle \int \frac{2x}{e^x}dx = \int 2xe^{-x}dx $ $\displaystyle = 2 \bigg(\bigg[ -xe^{-x}\bigg] - \int -e^{-x} dx \bigg) $
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