stuck on integration!

[SUhottie10]

i dont know how to put an integral sign so im gonna use | as the integral sign

|x^3 sqrt 2+x^4 dx = 1/4 | sqrt u du

u = 2+x^4 v = sqrt u
du = 4x^3dx dv = 1/2 u^-1/2

I cant figure out how to get dv in the integral.... if someone can help itd be great! thanks!

 

[daon]

Neglecting the little "fixes" involving constants, you essentially have an integral of the form \(\displaystyle \L \int u^{\frac{1}{2}} du\).

 

[SUhottie10]

yes i understand that but i dont see how i will get a u^-1/2

 

[stapel]

yes i understand that but i dont see how i will get a u^-1/2

I think the point was that you have a simple u-substitution here; you don't need to use parts. If u = sqrt[2 + x⁴], then what is du?

Eliz.