# Weird integral

1. Dec 9, 2003

### tandoorichicken

How do I do this
$$\int x\sqrt[3]{x-1}\,dx$$
?

2. Dec 9, 2003

Seems like a good candidate for integration by parts.

Let u=x and dv=(x-1)^1/3

3. Dec 9, 2003

### gnome

I assume that's x3 * &radic;(x-1) [ and not x * (x-1)^(1/3) ].

You can use integration by parts (3 times).

(There may be an easier way, but I don't see it.)

4. Dec 10, 2003

### himanshu121

U can substitute $$x-1=t^2$$

5. Dec 10, 2003

### himanshu121

U can substitute $$x-1=t^2$$

6. Dec 10, 2003

### tandoorichicken

Actually it is x(x-1)^(1/3), So I hope all this works for my problem. Actually, with whatever I've learned so far my teach says it can't be done, I was just curious if it actually can.

7. Dec 10, 2003

### HallsofIvy

Staff Emeritus
Let u= x-1. Then x= u+1 and dx= du so the integrand is

(u+1)u1/3du= (u4/3+ u1/3)du and it's easy.