Solving the \int x\sqrt[3]{x-1}\,dx Integration Problem

[tandoorichicken]

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

 

[LeBrad]

Seems like a good candidate for integration by parts.

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

 

[gnome]

I assume that's x³ * √(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.)

 

[himanshu121]

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

 

[himanshu121]

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

 

[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.

 

[HallsofIvy]

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

(u+1)u^1/3du= (u^4/3+ u^1/3)du and it's easy.