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- Thread starter Yuuki
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- Jan 26, 2012

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I would definitely recommend $w=1+2x$ to begin. Linear substitutions like that cost you nothing, and could gain you quite a bit, as in this case. What is the resulting $w$ integral?

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- Jan 26, 2012

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Here, let me type it out $\LaTeX$ style:

$$\frac{1}{4e} \int \frac{e^{w}(w-1)}{w^{2}} \, dw.$$

Is that what you have?

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- Jan 26, 2012

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Another way to solve this integral is to notice that

\begin{align*}

\frac{d}{dx} \frac{f(x)}{g(x)}&=\frac{g(x)\, f'(x)-f(x) \, g'(x)}{(g(x))^{2}} \\

\frac{f(x)}{g(x)}+C&=\int \frac{g(x)\, f'(x)-f(x) \, g'(x)}{(g(x))^{2}} \, dx.

\end{align*}

The quotient rule doesn't come in handy all that often, but when it does, it surprises you.