1. ## Integral of [e^(x^(1/b))]dx

I never learned how to integrate this type of function. Can anyone offer insight into how to integrate functions of the form

e^(x^(1/b)) with respect to x

and similar functions?

Thanks

2. I do not believe this has a closed form. That is probably why you are having trouble.

3. What do you mean by closed form?

Also, maybe I should be more specific.

How do you integrate e^(x^(1/4)) for example?

4. $
\int {e^{x^{\frac{1}{b}} } } dx = \int {e^{\frac{x}{b}} } dx = be^{\frac{x}{b}} + c
$

5. Originally Posted by Bastian
$
\int {e^{x^{\frac{1}{b}} } } dx = \int {e^{\frac{x}{b}} } dx = be^{\frac{x}{b}} + c
$
Sorry but this 'solution' is complete nonsense. $x^{1/b} \neq \frac{x}{b}$.

6. Originally Posted by imabookie3
What do you mean by closed form?

Also, maybe I should be more specific.

How do you integrate e^(x^(1/4)) for example?
I don't have time now but it can be done (as can your general case in post #1). See attachment.

7. Oh, I see, I misread. A general form may be tricky, but for $\int e^{x^{\frac{1}{4}}} dx$

We can use parts.

Let $u=4x^{\frac{3}{4}}, \;\ v=e^{x^{\frac{1}{4}}}, \;\ du=3x^{\frac{-1}{4}}dx, \;\ dv=\frac{1}{4}e^{x^{\frac{1}{4}}}x^{\frac{-3}{4}}dx$

Then we can put it together and get:

$4x^{\frac{3}{4}}e^{x^{\frac{1}{4}}}-3\int\frac{e^{x^{\frac{1}{4}}}}{x^{\frac{1}{4}}}dx$

Now, try applying parts again. Can you do that?.

You are shooting for this when you're done:

$4x^{\frac{3}{4}}e^{x^{\frac{1}{4}}}-12x^{\frac{1}{2}}e^{x^{\frac{1}{4}}}+24x^{\frac{1} {4}}e^{x^{\frac{1}{4}}}-24e^{x^{\frac{1}{4}}}$.

You can factor out the e term to tidy up.

There is a pattern to these which may lead us to a general form.

$\int e^{x^{\frac{1}{4}}}dx=4e^{x^{\frac{1}{4}}}\left(x^ {\frac{3}{4}}-3x^{\frac{2}{4}}+6x^{\frac{1}{4}}-6\right)$

$\int e^{x^{\frac{1}{5}}}dx=5e^{x^{\frac{1}{5}}}\left(x^ {\frac{4}{5}}-4x^{\frac{3}{5}}+12x^{\frac{2}{5}}-24x^{\frac{1}{5}}+24\right)$

$\int e^{x^{\frac{1}{6}}}dx=6e^{x^{\frac{1}{6}}}\left(x^ {\frac{5}{6}}-5x^{\frac{4}{6}}+20x^{\frac{3}{6}}-60x^{\frac{2}{6}}+120x^{\frac{1}{6}}-120\right)$

See a pattern you can derive a general form from?. I think so.

For instance, one way could be $be^{x^{\frac{1}{b}}}\left(x^{\frac{b-1}{b}}+(b-1)x^{\frac{b-2}{b}}-(b-2)(b-1)x^{\frac{b-3}{b}}+(b-3)(b-2)(b-1)x^{\frac{b-4}{b}}-....\right)$

There is a start for you.

8. Originally Posted by galactus
Oh, I see, I misread. A general form may be tricky, but for $\int e^{x^{\frac{1}{4}}} dx$

We can use parts.

Let $u=4x^{\frac{3}{4}}, \;\ v=e^{x^{\frac{1}{4}}}, \;\ du=3x^{\frac{-1}{4}}dx, \;\ dv=\frac{1}{4}e^{x^{\frac{1}{4}}}x^{\frac{-3}{4}}dx$

Then we can put it together and get:

$4x^{\frac{3}{4}}e^{x^{\frac{1}{4}}}-3\int\frac{e^{x^{\frac{1}{4}}}}{x^{\frac{1}{4}}}dx$

Now, try applying parts again. Can you do that?.

You are shooting for this when you're done:

$4x^{\frac{3}{4}}e^{x^{\frac{1}{4}}}-12x^{\frac{1}{2}}e^{x^{\frac{1}{4}}}+24x^{\frac{1} {4}}e^{x^{\frac{1}{4}}}-24e^{x^{\frac{1}{4}}}$.

You can factor out the e term to tidy up.

There is a pattern to these which may lead us to a general form.

$\int e^{x^{\frac{1}{4}}}dx=4e^{x^{\frac{1}{4}}}\left(x^ {\frac{3}{4}}-3x^{\frac{2}{4}}+6x^{\frac{1}{4}}-6\right)$

$\int e^{x^{\frac{1}{5}}}dx=5e^{x^{\frac{1}{5}}}\left(x^ {\frac{4}{5}}-4x^{\frac{3}{5}}+12x^{\frac{2}{5}}-24x^{\frac{1}{5}}+24\right)$

$\int e^{x^{\frac{1}{6}}}dx=6e^{x^{\frac{1}{6}}}\left(x^ {\frac{5}{6}}-5x^{\frac{4}{6}}+20x^{\frac{3}{6}}-60x^{\frac{2}{6}}+120x^{\frac{1}{6}}-120\right)$

See a pattern you can derive a general form from?. I think so.

For instance, one way could be $be^{x^{\frac{1}{b}}}\left(x^{\frac{b-1}{b}}+(b-1)x^{\frac{b-2}{b}}-(b-2)(b-1)x^{\frac{b-3}{b}}+(b-3)(b-2)(b-1)x^{\frac{b-4}{b}}-....\right)$

There is a start for you.
Maybe its been too long but

Let $\varphi=x^{\frac{1}{4}}\Rightarrow{\varphi^4=x}$

$\therefore{4\varphi^3{d\varphi}=dx}$

$\int{e^{x^{\frac{1}{4}}}dx\overbrace{{\mapsto}}^{\ varphi=x^\frac{1}{4}}}\int{e^{\varphi}\cdot{4\varp hi^3}d\varphi}$
An easy problem?

and then we can see from that

$\int{e^{x^{\frac{1}{b}}dx}\overbrace{\mapsto}^{\va rphi=x^{\frac{1}{b}}}\int{e^{\varphi}}\cdot{b\varp hi^{b-1}}d\varphi}\quad\forall{b}\in\mathbb{Z^+}$

9. Originally Posted by Mathstud28
Maybe its been too long but

Let $\varphi=x^{\frac{1}{4}}\Rightarrow{\varphi^4=x}$

$\therefore{4\varphi^3{d\varphi}=dx}$

$\int{e^{x^{\frac{1}{4}}}dx\overbrace{{\mapsto}}^{\ varphi=x^\frac{1}{4}}}\int{e^{\varphi}\cdot{4\varp hi^3}d\varphi}$
An easy problem?

and then we can see from that

$\int{e^{x^{\frac{1}{b}}dx}\overbrace{\mapsto}^{\va rphi=x^{\frac{1}{b}}}\int{e^{\varphi}}\cdot{b\varp hi^{b-1}}d\varphi}\quad\forall{b}\in\mathbb{Z^+}$
Now I have time, I come back to say these very things and see I'm beaten to the punch. *Socko*

10. Originally Posted by mr fantastic
Now I have time, I come back to say these very things and see I'm beaten to the punch. *Socko*
I feel the love

11. Wow, look who's back from his long hiatus.

I am referring to you, Mathstud. Where ya' been?.

12. Originally Posted by galactus
Wow, look who's back from his long hiatus.

I am referring to you, Mathstud. Where ya' been?.
School, MMA, Girls

13. Originally Posted by mr fantastic
Sorry but this 'solution' is complete nonsense. $x^{1/b} \neq \frac{x}{b}$.
ohh you're right, lapsus