How to integrate this

Integral of : 3*pi*x^(2/3) * [ ( x^(2/3) + 1 ) / x^(2/3) ]^(1/2)

Thanks in advance,

Note that:
∫ 3πx^(2/3)√{[x^(2/3) + 1]/x^(2/3)} dx
= 3π ∫ x^(2/3)√[x^(2/3) + 1]/x^(1/3) dx.

At this point, we can substitute:
u = x^(2/3) + 1 ==> x^(2/3) = u - 1 and du = (2/3)x^(-1/3) = 2/[3x^(1/3)] dx.

This gives
3π ∫ x^(2/3)√[x^(2/3) + 1]/x^(1/3) dx
= 9π/2 ∫ x^(2/3)√[x^(2/3) + 1] {2/[3x^(1/3)] dx}
= 9π/2 ∫ (u - 1)√u du, by applying substitutions
= 9π/2 ∫ [u^(3/2) - √u] du
= (9π/2)[(2/5)u^(5/2) - (2/3)u^(3/2)] + C, by integrating
= (9π/5)u^(5/2) - 3π*u^(3/2) + C
= (9π/5)[x^(2/3) + 1]^(5/2) - 3π*[x^(2/3) + 1]^(3/2) + C, by back-substituting.

I hope this helps!

∫ 3 * π * x^(2/3) * [ ( x^(2/3) + 1 ) / x^(2/3) ]^(1/2) dx

∫ 3 * π * x^(2/3) * [ [ ( x^(2/3) + 1 ) ]^(1/2) / [ x^(2/3) ]^(1/2) ] dx

∫ 3 * π * [ x^(2/3) ]^(1/2) ] * [ [ ( x^(2/3) + 1 ) ]^(1/2) dx

∫ 3 * π * [ x^( (2/3) * (1/2) ) ] * [ [ ( x^(2/3) + 1 ) ]^(1/2) dx

∫ 3 * π * [ x^(1/3) ] * [ ( x^(2/3) + 1 ) ]^(1/2) dx

u = x^(2/3) + 1 =====> u - 1 = x^(2/3)

du = (2/3) * x^(-1/3) dx =====> du = (2/3) * (1 / x^(1/3)) dx ====> (3/2) * x^(1/3)du =dx


∫ 3 * π * [ x^(1/3) ] * [ u ]^(1/2) * (3/2) * x^(1/3) * du


∫ (9/2) * π * [ x^(2/3) ] * [ u ]^(1/2) * du

∫ (9/2) * π * ( u - 1 ) * √(u) * du

∫ (9/2) * π * ( u^(3/2) - √(u) ) * du

(9/2) * π * ( u^(5/2) / (5/2) ) - ( (u)^(3/2) / (3/2) ) + C

(9/2) * π * ( (2/5) * u^(5/2) ) - ( (2/3) * (u)^(3/2) ) + C

π * ( (9/5) * u^(5/2) ) - ( (9/3) * (u)^(3/2) ) + C

π * ( (9/5) * u^(5/2) ) - ( 3 * (u)^(3/2) ) + C

π * ( (9/5) * [ x^(2/3) + 1 ]^(5/2) ) - ( 3 * [ x^(2/3) + 1 ]^(3/2) ) + C

(9/5) * π * [ x^(2/3) + 1 ]^(5/2) - 3 * π * [ x^(2/3) + 1 ]^(3/2) + C