Can someone please help with this calculus problem?

Question

integral of [x^5.(x^3 + 1)^(1/3)].dx

Answer 1

Integ [x^5.(x^3 + 1)^(1/3)] dx

Well your choices are Substitutions and/or Integration by Parts.

(x^2 + 1) or ^(1/2) would suggest a trigonometric/hyperbolic substitution, but (x^3 + 1)^(1/3) doesn't.

(x^3 + 1)^(1/3) is going to be nasty,
so try substitution w=x^3 + 1 , dw=3x^2dx
Noting x^5 = 1/3 * x^3 * 3x^2 ...

Then Integ [x^5.(x^3 + 1)^(1/3)] dx
= 1/3 Integ [x^3.(x^3 + 1)^(1/3) *3x^2 dx]
= 1/3 Integ [(w-1) (w)^(1/3) *dw]
= 1/3 Integ [w^(4/3) - w^(1/3)] dw
= 1/3 (3/7 w^(7/3) - 3/4 w^(4/3)) + C
= w^(4/3) * (1/7 w - 1/4) + C

These things are like an open-ended rope, you can spin yourself into trouble.
The question I always ask myself to check is: did my approach reduce the complexity and/or the order of the expression, or else translate it to some known standard form for which an identity exists? If not, then the step was no good.

In this case you really want to eliminate the (x^3 + 1)^(1/3) term.

Answer 2

integral of [x^5.(x^3 + 1)^(1/3)].dx = I (Say)
Put y = x^3, then dy =3x^2 dx. Then the above integral becomes:
I = integral of [ (1/3) y (y+1)^(1/3)] dy.
=(1/3) integral of [y(y+1)^(1/3)]dy
=(1/3) integral of [(y+1 -1)(y+1)^(1/3)] dy
=(1/3) integral of [(y+1)^(4/3) -(y+1)^(1/3)] dy
= (1/3)[ (3/7) (y+1)^(7/3) - (3/4) (y+1)^(4/3)] + C(constant)
= (1/28)(y+1)^(4/3) [4y+4 - 7] +C
=(1/28) (4y -3) (y+1)^(4/3) + C. OR
I = (1/28) (4x^3 -3) (x^3 +1)^(4/3) + C., putting back y = x^3.

Verification:
Derivative of this
= (1/28)[(4x^3 - 3) (4/3)(x^3 +1)^(1/3) .3x^2 + 12x^2(x^3+1)^(4/3)]
= (1/7)x^2 (x^3 +1) ^(1/3) [4x^3 - 3 + 3(x^3 +1)]
= x^5 (x^3 +1)^(1/3) OOKKKK.

Answer 3

Integrate {(x^5)*(x³ + 1)^(1/3)} with respect to x.

∫{(x^5)*(x³ + 1)^(1/3)}
= (1/3)∫{(3x²)(x³)(x³ + 1)^(1/3)}dx

Let
u = x³ + 1
du = 3x² dx
u - 1 = x³

= (1/3)∫{(u - 1)[u^(1/3)]}du = (1/3)∫{u^(4/3) - u^(1/3)}du
= (1/3){(3/7)u^(7/3) - (3/4)u^(4/3) + C
= (1/7)u^(7/3) - (1/4)u^(4/3) + C
= {4u^(7/3) - 7u^(4/3)} / 28 + C
= {4(x³ + 1)^(7/3) - 7(x³ + 1)^(4/3)} / 28 + C
= (x³ + 1)^(4/3){4(x³ + 1) - 7} / 28 + C
= (x³ + 1)^(4/3){4x³ - 3} / 28 + C