What is the integral of ((x-1))/(x^{2/3)+x^{1/3}+1} ?

The integral of ((x-1))/(x^{2/3)+x^{1/3}+1} is 3/4 x^{4/3}-x+C

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integral of ((x-1))/(x^{2/3)+x^{1/3}+1}

\int \frac{( x - 1 )}{x ^{\frac{2}{3}} + x ^{\frac{1}{3}} + 1} d x

\frac{3}{4} x^{\frac{4}{3}} - x + C

Solution steps

\int \frac{x - 1}{x ^{\frac{2}{3}} + x ^{\frac{1}{3}} + 1} d x

Apply u-substitution

: \int 3 u^{2} (u - 1) d u

= \int 3 u^{2} (u - 1) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int u^{2} (u - 1) d u

Expand

u^{2} (u - 1) : u^{3} - u^{2}

= 3 \cdot \int u^{3} - u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int u^{3} d u - \int u^{2} d u)

\int u^{3} d u = \frac{u ^{4}}{4}

\int u^{2} d u = \frac{u ^{3}}{3}

= 3 (\frac{u ^{4}}{4} - \frac{u ^{3}}{3})

Substitute back

u = x^{\frac{1}{3}}

= 3 \frac{( x ^{\frac{1}{3}} ) ^{4}}{4} - \frac{( x ^{\frac{1}{3}} ) ^{3}}{3}

Simplify

3 \frac{( x ^{\frac{1}{3}} ) ^{4}}{4} - \frac{( x ^{\frac{1}{3}} ) ^{3}}{3} : \frac{3}{4} x^{\frac{4}{3}} - x

= \frac{3}{4} x^{\frac{4}{3}} - x

Add a constant to the solution

= \frac{3}{4} x^{\frac{4}{3}} - x + C

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What is the integral of ((x-1))/(x^{2/3)+x^{1/3}+1} ?
The integral of ((x-1))/(x^{2/3)+x^{1/3}+1} is 3/4 x^{4/3}-x+C