# `int root(3)(x)/(root(3)(x) - 1) dx` Find the indefinite integral by u substitution. (let u be the denominator of the integral)

Solving indefinite integral by u-substitution, we follow:

`int f(g(x))*g'(x) = int f(u) *du` where we let `u = g(x)` .

By following the instruction to let "u" be the denominator  of the integral,

it means we let: u = `root(3)(x) -1`

Find the derivative of "u" which is` du = 1/(3x^(2/3))dx`

Then `du =1/(3x^(2/3))dx` can be rearrange into `3x^(2/3)du =dx` .

Applying u-substitution using `u =root(3)(x)-1` and `3x^(2/3)du =dx` .

`int root(3)(x)/(root(3)(x)-1) dx = int root(3)(x)/u*3x^(2/3)du`

`= int (x^(1/3)*3x^(2/3))/udu`

` =int (3x^(1/3+2/3))/udu`

` =3 int x/udu`

Note: `x^(1/3+2/3) = x^(3/3)`

`=x^1 or x`

Algebraic techniques:

From` u = root(3)(x)-1` , we can rearrange it into `root(3)(x)=u+1` .

Raising both sides by a power 3:

`(root(3)(x))^3 =(u+1)^3`

`x = (u+1)*(u+1)*(u+1)`

By FOIL: `(u+1)*(u+1) = u*u +u*1+1*u+1*1`

`= u^2+u+u+1`

`= u^2+2u+1`

Then let `(u+1)(u+1) = u^2 +2u +1 in (u+1)(u+1)(u+1)` :

`(u+1)(u+1)(u+1) = (u+1)*(u^2+2u+1)`

Applying distributive property:

`(u+1)(u^2+2u+1) = u *(u^2+2u+1) + 1*(u^2+2u+1)`

`= u^3 +2u^2+u +u^2+2u+1`

`=u^3+3u^2+3u+1`

then `x = (u+1)*(u+1)*(u+1)` is the same as

`x =u^3+3u^2+3u+1`

Substitute` x=u^3+3u^2+3u+1 in 3 int x/udu` :

`3 int x/udu = 3 int (u^3+3u^2+3u+1 )/u du`

` = 3int (u^3/u+(3u^2)/u+(3u)/u+1/u) du`

`=3int (u^2+3u+3+1/u) du`

Evaluating each term in separate integral:

`3 * [ int u^2 *du+ int 3u*du+int 3*du+ int 1/u du]`

where:

`int u^2 *du = u^3/3`

`int 3u*du =(3u^2)/2`

`int 3*du = 3u`

`int 1/u du= ln|u|`

`3 * [ int u^2 *du+ int 3u*du+int 3*du+ int 1/u du]` becomes:

`3*[u^3/3 +(3u^2)/2 +3u+ln|u|] +C= 3u^3/3 +(9u^2)/2 +9u+3ln|u|+C`

Substitute u = root(3)(x)-1:

`3u^3/3 +(9u^2)/2 +9u+3ln|u| +C = (root(3)(x)-1)^3 +(9(root(3)(x)-1)^2)/2 +9(root(3)(x)-1)+3ln|(root(3)(x)-1)| +C`

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