exponential integral

[palangi]

I'm trying to find the indefinite integral of e^(x^(1/3))dx.

I wanted it to be 3e^(x^(1/3)), but I think that it isn't because the exponent also has an exponent, is this correct?

If it is, I'm thinking u substitution (u=x^(1/3), -2/3x^(-2/3)du=dx) so -2/3 Integral of e^u * x^(-2/3)du and then integration by parts, but I'm not getting anywhere that looks like a final answer.

Help please

 

[galactus]

Re: exponetial integral

You can let \(\displaystyle x=u^{3}, \;\ dx=3u^{2}du\)

\(\displaystyle 3\int u^{2}e^{u}du\)

Now, parts will do.

Or, we can try a method perhaps you have not seen.

It works well for integrating the product of as polynomial and e.

\(\displaystyle \frac{d}{du}p(u)e^^{u}=u^{2}e^{u}\)

Let \(\displaystyle p(u)=au^{2}+bu+c\)

\(\displaystyle u^{2}+bu+c=u^{2}e^{u}\)

\(\displaystyle (2au+b)e^{u}+(au^{2}+bu+c)e^{u}=u^{2}e^{u}\)

\(\displaystyle au^{2}+(2a-b)u+(b+c))e^{u}=u^{2}e^{u}\)

Equate coefficients:

\(\displaystyle a=1, \;\ 2a+b=0, \;\ b+c=0\)

\(\displaystyle a=1, \;\ b=-2, \;\ c=2\)

So, we finally get:

\(\displaystyle \frac{d}{du}(u^{2}-2u+2)e^{u}=u^{2}e^{u}\)

Integrate both sides to cancel the derivative:

\(\displaystyle (u^{2}-2u+2)e^{u}=\int u^{2}e^{u}du\)

Resub:

\(\displaystyle 3e^{x^{\frac{1}{3}}}(x^{\frac{2}{3}}-2x^{\frac{1}{3}}+2)\)

 

[palangi]

The first part:

I'm sorry, I don't know why this is not making sense...I don't understand why you solve for x=u^3 instead of u=x^(1/3)

I do understand that it is the same thing, but because I'm just learning these it doesn't make sense to me that we are solving for dx and not du. And if we do it like that, shouldn't both ways work? But I can't get u=x^(1/3) du=1/3x^(-2/3) to integrate, which is probably why you solved for x instead of u, but I'm not understanding how we make that jump.

Am I making sense? I'm frustrated with this problem , thank you for your patience.

 

[galactus]

it doesn't make sense to me that we are solving for dx and not du

After making the u substitution, you have to have something in terms of du to replace dx with.

After you make your subs, there can no longer be any x's. It has to be entirely in terms of u.


If you are trying parts from the beginning, then

\(\displaystyle \int e^{x^{\frac{1}{3}}}dx\)

Let \(\displaystyle u=e^{x^{\frac{1}{3}}}, \;\ du=\frac{e^{x^{\frac{1}{3}}}}{3x^{\frac{2}{3}}}dx, \;\ dv=dx, \;\ v=x\)

\(\displaystyle uv-\int vdu\)

\(\displaystyle xe^{x^{\frac{1}{3}}}-\frac{1}{3}\int\frac{xe^{x^{\frac{1}{3}}}}{x^{\frac{2}{3}}}dx\)

Now, using a substitution, we can let \(\displaystyle u=x^{\frac{1}{3}}, \;\ du=\frac{x^{\frac{1}{3}}}{3x^{\frac{2}{3}}}dx\)

You can use a sub now or continue with another application of parts.

There are various ways to go about it. Personally, I prefer the method I outlined first.

 

[palangi]

Thank you for your help. I have a test in two hours so I'll stick with the substitution and by parts for now, but will go try and learn your preferred method after, any time I can get by without using fractions, I'm in. Thanks again.

 

[galactus]

As a rule:

\(\displaystyle \int u^{n}e^{u}du=u^{n}e^{u}-n\int u^{n-1}e^{u}du\)