Author Topic: anti-differentiation question (Read 670 times) Tweet Share

monokekie · « **on:** August 02, 2009, 10:05:33 am »

Thank you!

Collin Li · « **Reply #1 on:** August 02, 2009, 10:31:50 am »

$u = 1 + e^x$

and $e^{2x} = e^x \cdot e^x$ so the integral is: $\frac{u-1}{u}\cdot du$

(Elaborating: one of the $e^x$ transforms the $dx$ into $du$ . The other is $u - 1 = 1 + e^x - 1 = e^x$ )

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Author Topic: anti-differentiation question (Read 670 times) Tweet Share

monokekie

anti-differentiation question

Collin Li

Re: anti-differentiation question

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