Integration

[renyikouniao]

Question:Integrate (1-e^(2x))/e^x dx
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)

 

[Jameson]

Question:Integrate (1-e^(2x))/e^x dx
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)

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?

 

[renyikouniao]

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

 

[Prove It]

Question:Integrate (1-e^(2x))/e^x dx
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)

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

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

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

 

[MarkFL]

Another approach is to rewrite the integrand as follows:

\(\displaystyle \frac{1-e^{2x}}{e^x}\cdot\frac{e^{-x}}{e^{-x}}=e^{-x}-e^x=-2\sinh(x)\)

and now we have:

\(\displaystyle -2\int\sinh(x)\,dx=-2\cosh(x)+C=-e^x-e^{-x}+C\)