# Integral questions

1. Feb 7, 2004

### stoffer

Can anybody help me with the integral of 1/ (x^1/3 + x^1/4)
(cube root and fourth root of x) I dont 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 havent learned yet)

2. Feb 7, 2004

### 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...

3. Feb 7, 2004

### 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)

4. Feb 7, 2004

### stoffer

ok thanks for your help guys

5. Mar 14, 2007

### wesywes

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

6. Mar 22, 2007

### d_leet

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

7. Mar 22, 2007

### d_leet

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

8. Mar 22, 2007

### 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:

$\frac{x^{12}}{x^4 + x^3} = \frac{x^9}{1 + x} = \frac{x^9 + 1 - 1}{1+x}$

Then proceed...

9. Mar 23, 2007

### dextercioby

Actually it's a power less in the numerator:

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

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

10. Mar 23, 2007

### morphism

I didn't use the substitution...