Ask your own question, for FREE!
Mathematics 6 Online
OpenStudy (shamil98):

∫xe^x^3 dx integration help

OpenStudy (shamil98):

\[\huge \int\limits_{}^{} xe^{x^3} dx\]

OpenStudy (shamil98):

u = x^3 du = 3x^2 dx ?

OpenStudy (shamil98):

what would i write xdx as ?

OpenStudy (shamil98):

\[\huge xdx = \frac{ 1 }{ 3 } du\] ?

OpenStudy (isaiah.feynman):

That's not integration by parts, that's u substitution you are doing

OpenStudy (shamil98):

oh... new to this part of integration x_x

zepdrix (zepdrix):

Hmm this doesn't look solvable using normal methods. Are you sure you posted the question correctly? Have you learned about the `gamma function`? Or is this just for a calculus class?

OpenStudy (shamil98):

it's for calc 2.

OpenStudy (schrodingers_cat):

Perhaps a power series solution?

OpenStudy (anonymous):

Sham is this question in a book?

OpenStudy (shamil98):

nope.

OpenStudy (anonymous):

Make it up?

OpenStudy (shamil98):

yep

OpenStudy (anonymous):

Go away

OpenStudy (shamil98):

noo i saw xe^x^2 why is it soo hard to do xe^x^3

OpenStudy (schrodingers_cat):

The problem is integrating e^x^3 if you won't do do integration by parts.

OpenStudy (schrodingers_cat):

want to*

OpenStudy (shamil98):

would substitution work then?

OpenStudy (schrodingers_cat):

No, because you would need a factor of x^2 in front if you wanted to undo the chain rule.

OpenStudy (schrodingers_cat):

and or do a substitution.

OpenStudy (schrodingers_cat):

I do not know if you have done power series yet but a power series representation would be x + x^4 + x^7/2 + x^10/6 + x^13/24 You could integrate from there and see what you get as a series.

OpenStudy (isaiah.feynman):

:P

zepdrix (zepdrix):

Ahhh I wish you would state that when you post the question! :( `That you made it up`. Integrals aren't nice and friendly like derivatives: you can't just make something up and expect it to have a simple solution. If you had started with something instead like:\[\Large \int\limits x^3e^{x^3}\;dx\]Then yes you could do `by parts` on that.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!