How Do You Integrate Functions Involving Powers and Trigonometric Terms?

[Blade]

INT x^(1/3) / x^(1/3) -1 dx
u = x^(1/3)
x = u^3
dx = 3u^2du

INT (u/u-1)3u^2du = 3
INT u^3/(u-1) du
INT [u^2 + u + 1/(u-1)]du
Then just integrate each part?

-----------
Find the area of the region bounded.

y=2x - tan(0.3x), x-1, x=4, y=0
Need some guidance where to start here, and how the conditions will be used.

 

[HallsofIvy]

Looks reasonable to me!

What guidance do you need for the second problem?

Just integrate 2x- tan(0.3x) from 1 to 4.

 

[PrudensOptimus]

Originally posted by Blade
INT x^(1/3) / x^(1/3) -1 dx
u = x^(1/3)
x = u^3
dx = 3u^2du

INT (u/u-1)3u^2du = 3
INT u^3/(u-1) du
INT [u^2 + u + 1/(u-1)]du
Then just integrate each part?

-----------
Find the area of the region bounded.

y=2x - tan(0.3x), x-1, x=4, y=0
Need some guidance where to start here, and how the conditions will be used.

I don't see any ln in the integrand.