Originally Posted by
galactus Oh, I see, I misread. A general form may be tricky, but for $\displaystyle \int e^{x^{\frac{1}{4}}} dx$
We can use parts.
Let $\displaystyle u=4x^{\frac{3}{4}}, \;\ v=e^{x^{\frac{1}{4}}}, \;\ du=3x^{\frac{-1}{4}}dx, \;\ dv=\frac{1}{4}e^{x^{\frac{1}{4}}}x^{\frac{-3}{4}}dx$
Then we can put it together and get:
$\displaystyle 4x^{\frac{3}{4}}e^{x^{\frac{1}{4}}}-3\int\frac{e^{x^{\frac{1}{4}}}}{x^{\frac{1}{4}}}dx$
Now, try applying parts again. Can you do that?.
You are shooting for this when you're done:
$\displaystyle 4x^{\frac{3}{4}}e^{x^{\frac{1}{4}}}-12x^{\frac{1}{2}}e^{x^{\frac{1}{4}}}+24x^{\frac{1} {4}}e^{x^{\frac{1}{4}}}-24e^{x^{\frac{1}{4}}}$.
You can factor out the e term to tidy up.
There is a pattern to these which may lead us to a general form.
$\displaystyle \int e^{x^{\frac{1}{4}}}dx=4e^{x^{\frac{1}{4}}}\left(x^ {\frac{3}{4}}-3x^{\frac{2}{4}}+6x^{\frac{1}{4}}-6\right)$
$\displaystyle \int e^{x^{\frac{1}{5}}}dx=5e^{x^{\frac{1}{5}}}\left(x^ {\frac{4}{5}}-4x^{\frac{3}{5}}+12x^{\frac{2}{5}}-24x^{\frac{1}{5}}+24\right)$
$\displaystyle \int e^{x^{\frac{1}{6}}}dx=6e^{x^{\frac{1}{6}}}\left(x^ {\frac{5}{6}}-5x^{\frac{4}{6}}+20x^{\frac{3}{6}}-60x^{\frac{2}{6}}+120x^{\frac{1}{6}}-120\right)$
See a pattern you can derive a general form from?. I think so.
For instance, one way could be $\displaystyle be^{x^{\frac{1}{b}}}\left(x^{\frac{b-1}{b}}+(b-1)x^{\frac{b-2}{b}}-(b-2)(b-1)x^{\frac{b-3}{b}}+(b-3)(b-2)(b-1)x^{\frac{b-4}{b}}-....\right)$
There is a start for you.