Three indefinite integrals involving e^x

[rainyrainy906]

Thank you :)

View attachment 3775

 

[MarkFL]

I have moved this thread as it involved integral calculus, and I have given it a descriptive title so that people looking at the thread listing can see at a glance the nature of the questions being asked. For future reference we also ask that no more than two questions be posted in a thread so that it does not potentially become convoluted and hard to follow.

In order for us to be able to help you, we need to see what you have tried so we know where you are stuck. Can you post your work so far?

 

[rainyrainy906]

I have moved this thread as it involved integral calculus, and I have given it a descriptive title so that people looking at the thread listing can see at a glance the nature of the questions being asked. For future reference we also ask that no more than two questions be posted in a thread so that it does not potentially become convoluted and hard to follow.

In order for us to be able to help you, we need to see what you have tried so we know where you are stuck. Can you post your work so far?

Thank you a lot :)
Next time I'll pay more attention

 

[MarkFL]

I suspect there is a typo in the first problem, as it involves a hypergeometric anti-derivative, so let's look at the second one:

\(\displaystyle I=\int\frac{\left(x^2+x\right)e^x}{x+e^{-x}}\,dx\)

My first thought here is to multiply the integrand by \(\displaystyle 1=\frac{e^x}{e^x}\) to get:

\(\displaystyle I=\int\frac{\left(x^2+x\right)e^{2x}}{xe^x+1}\,dx\)

Now, if we look at the denominator, and see that by differentiation, we obtain:

\(\displaystyle \frac{d}{dx}\left(xe^x+1\right)=xe^x+e^x=e^x(x+1)\)

And then observe that we may write our integral as:

\(\displaystyle I=\int\frac{e^x\left(x+1\right)xe^{x}}{xe^x+1}\,dx\)

Now, I think we are ready to try integration by parts...can you see what substitutions you should try?

 

[Prove It]

I suspect there is a typo in the first problem, as it involves a hypergeometric anti-derivative, so let's look at the second one:

\(\displaystyle I=\int\frac{\left(x^2+x\right)e^x}{x+e^{-x}}\,dx\)

My first thought here is to multiply the integrand by \(\displaystyle 1=\frac{e^x}{e^x}\) to get:

\(\displaystyle I=\int\frac{\left(x^2+x\right)e^{2x}}{xe^x+1}\,dx\)

Now, if we look at the denominator, and see that by differentiation, we obtain:

\(\displaystyle \frac{d}{dx}\left(xe^x+1\right)=xe^x+e^x=e^x(x+1)\)

And then observe that we may write our integral as:

\(\displaystyle I=\int\frac{e^x\left(x+1\right)xe^{x}}{xe^x+1}\,dx\)

Now, I think we are ready to try integration by parts...can you see what substitutions you should try?

Integration by parts? Really?

$\displaystyle \begin{align*} \int{ \frac{\mathrm{e}^x\,\left( x + 1 \right) \, x\,\mathrm{e}^x}{x\,\mathrm{e}^x + 1} \,\mathrm{d}x} \end{align*}$

Let $\displaystyle \begin{align*} u = x\,\mathrm{e}^x + 1 \implies \mathrm{d}u = \mathrm{e}^x \, \left( x + 1 \right) \,\mathrm{d}x \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int{ \frac{u - 1}{u}\,\mathrm{d}u } &= \int{ 1 - \frac{1}{u}\,\mathrm{d}u} \end{align*}$

I don't see why we would need it :P

 

[MarkFL]

Yes, I am guilty of thinking IBP was what we would need before I even started, and so I "painted the bullseye" around where my arrow landed...hehehe.

 

[MarkFL]

For the third one, I would divide the numerator and denominator of the integrand by $x$ to obtain:

\(\displaystyle I=\int\frac{e^x+\frac{1}{x}}{e^x+\ln(x)}\,dx\)

Now, do you see an appropriate $u$-substitution?