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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: monokekie on August 02, 2009, 10:05:33 am

Title: anti-differentiation question
Post by: monokekie on August 02, 2009, 10:05:33 am: (http://img185.imageshack.us/img185/3063/33371823.png)Thank you!
Title: Re: anti-differentiation question
Post by: Collin Li 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 $u - 1 = 1 + e^x - 1 = e^x$ )