Integral of 1/(x^1/3 + x^1/4): Can Anyone Help?

[stoffer]

Can anybody help me with the integral of 1/ (x^1/3 + x^1/4)
(cube root and fourth root of x) I don't really know where to start.

Also my roomate and i were wondering if x^sin(x) exists or if it has to be expressed and integrated as some sort of series.(something i haven't learned yet)

 

[Muzza]

Perhaps you could rewrite it as 1 / (x^(1/4) * (1 + x^(1/12))) and use partial fractions? I don't really know. Looks like it'll be pretty messy...

 

[matt grime]

try substituting x = y^12

as for x^sin(x).. it's exp{log(x^sinx)} = exp{sinx*log(x)}

that's the standard way of defining x^f(x)

 

[stoffer]

ok thanks for your help guys

 

[wesywes]

Can anybody help me with the integral of 1/ (x^1/3 + x^1/4)
(cube root and fourth root of x) I don't really know where to start.

Also my roomate and i were wondering if x^sin(x) exists or if it has to be expressed and integrated as some sort of series.(something i haven't learned yet)

just take the ln of that.. and do the rest intuitively, but hey what do i know, I am only 16.

 

[d_leet]

This is actually a very simple integral. Rewrite it as [tex] \int x^{-\frac{1}{3}} + x^{-\frac{1}{2}} dx [/tex]

That is certainly not the same as the function in the original post.

 

[d_leet]

Sorry, I meant [tex] \int x^{-\frac{1}{3}} + x^{-\frac{1}{4}} dx [/tex]

That still isn't the same function. You can't manipulate fractions like that, it just doesn't work.

 

[morphism]

This is why I'm not a fan of rushing people into calculus without a solid foundation in the basics of algebra. (In reference to d_leet's quotes.)

Anyway, matt's substitution kills this integral. You can also re-write the integrand as:

[itex]\frac{x^{12}}{x^4 + x^3} = \frac{x^9}{1 + x} = \frac{x^9 + 1 - 1}{1+x}[/itex]

Then proceed...

 

[dextercioby]

Anyway, matt's substitution kills this integral. You can also re-write the integrand as:

[itex]\frac{x^{12}}{x^4 + x^3} = \frac{x^9}{1 + x} = \frac{x^9 + 1 - 1}{1+x}[/itex]

Then proceed...

Actually it's a power less in the numerator:

[itex] I=\int \frac{dx}{\sqrt[3]{x}+\sqrt[4]{x}}=12 \int \frac{y^{8}}{y+1}{}dy [/itex]

,where [itex] y=\sqrt[12]{x} [/itex].

 

[morphism]

Actually it's a power less in the numerator:

[itex] I=\int \frac{dx}{\sqrt[3]{x}+\sqrt[4]{x}}=12 \int \frac{y^{8}}{y+1}{}dy [/itex]

,where [itex] y=\sqrt[12]{x} [/itex].

I didn't use the substitution...