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#### renyikouniao

##### Member

- Jun 1, 2013

- 41

My solution:=integrate 1/e^x-e^(2x)/e^x

=integrate e^(-x)-e^x

This is where i am stuck,I don't know how to integrate e^(-x)

- Thread starter renyikouniao
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- Thread starter
- #1

- Jun 1, 2013

- 41

My solution:=integrate 1/e^x-e^(2x)/e^x

=integrate e^(-x)-e^x

This is where i am stuck,I don't know how to integrate e^(-x)

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- #2

- Jan 26, 2012

- 4,040

Hi renyikouniao,

My solution:=integrate 1/e^x-e^(2x)/e^x

=integrate e^(-x)-e^x

This is where i am stuck,I don't know how to integrate e^(-x)

Let $u=-x$ and $du=-dx$. We can now transform this integral into \(\displaystyle \int -e^{u}du\) and then back-substitute. What do you get when you do that?

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- #3

- Jun 1, 2013

- 41

Hi renyikouniao,

Let $u=-x$ and $du=-dx$. We can now transform this integral into \(\displaystyle \int -e^{u}du\) and then back-substitute. What do you get when you do that?

That make sence!!Thank you very much

\(\displaystyle \displaystyle \begin{align*} \int{\frac{1 - e^{2x}}{e^x}\,dx} &= \int{\frac{e^x \left( 1 - e^{2x} \right) }{e^{2x}}\,dx} \end{align*}\)

My solution:=integrate 1/e^x-e^(2x)/e^x

=integrate e^(-x)-e^x

This is where i am stuck,I don't know how to integrate e^(-x)

Now make the substitution \(\displaystyle \displaystyle \begin{align*} u = e^x \implies du = e^x\,dx \end{align*}\) and the integral becomes

\(\displaystyle \displaystyle \begin{align*} \int{\frac{e^x\left( 1 - e^{2x} \right) }{e^{2x}}\,dx} &= \int{\frac{1 - u^2}{u^2}\,du} \\ &= \int{\frac{1}{u^2} - 1 \, du} \\ &= \int{u^{-2} - 1\,du} \\ &= \frac{u^{-1}}{-1} - u + C \\ &= -\left( e^x \right) ^{-1} - e^x + C \\ &= -e^{-x} - e^x + C \end{align*}\)

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