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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int (sec (4 x) - csc (\frac{2 x}{3})) d x

\int \frac{9 x ^{8} + 6}{( x ^{9} + 6 x ) ^{2}} d x

\int \frac{2 x y ( y + 2 )}{( 2 x ^{2} + y ^{2} ) ^{2}} d x

\int \frac{x ^{2} + 12}{x x ^{2} - 9} d x

\int \frac{1}{7 x - x ^{2}} d x

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