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

The integral of (x^2)/((1+2x)^{1/3)} is 1/320 (60x^2(1+2x)^{2/3}-36x(1+2x)^{2/3}+27(1+2x)^{2/3})+C

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

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

\frac{1}{320} (60 x^{2} (1 + 2 x)^{\frac{2}{3}} - 36 x (1 + 2 x)^{\frac{2}{3}} + 27 (1 + 2 x)^{\frac{2}{3}}) + C

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Apply u-substitution

: \int \frac{( u - 1 ) ^{2}}{8 u ^{\frac{1}{3}}} d u

= \int \frac{( u - 1 ) ^{2}}{8 u ^{\frac{1}{3}}} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{( u - 1 ) ^{2}}{u ^{\frac{1}{3}}} d u

Apply Integration By Parts

: u^{\frac{2}{3}} (u - 1)^{2} - \int \frac{5 u ^{2} - 4 u - 1}{3 u ^{\frac{1}{3}}} d u

= \frac{1}{8} (u^{\frac{2}{3}} (u - 1)^{2} - \int \frac{5 u ^{2} - 4 u - 1}{3 u ^{\frac{1}{3}}} d u)

\int \frac{5 u ^{2} - 4 u - 1}{3 u ^{\frac{1}{3}}} d u = \frac{1}{40} u^{\frac{2}{3}} (25 u^{2} - 32 u - 20)

= \frac{1}{8} (u^{\frac{2}{3}} (u - 1)^{2} - \frac{1}{40} u^{\frac{2}{3}} (25 u^{2} - 32 u - 20))

Substitute back

u = 1 + 2 x

= \frac{1}{8} ((1 + 2 x)^{\frac{2}{3}} (1 + 2 x - 1)^{2} - \frac{1}{40} (1 + 2 x)^{\frac{2}{3}} (25 (1 + 2 x)^{2} - 32 (1 + 2 x) - 20))

Simplify

\frac{1}{8} ((1 + 2 x)^{\frac{2}{3}} (1 + 2 x - 1)^{2} - \frac{1}{40} (1 + 2 x)^{\frac{2}{3}} (25 (1 + 2 x)^{2} - 32 (1 + 2 x) - 20)) : \frac{1}{320} (60 x^{2} (1 + 2 x)^{\frac{2}{3}} - 36 x (1 + 2 x)^{\frac{2}{3}} + 27 (1 + 2 x)^{\frac{2}{3}})

= \frac{1}{320} (60 x^{2} (1 + 2 x)^{\frac{2}{3}} - 36 x (1 + 2 x)^{\frac{2}{3}} + 27 (1 + 2 x)^{\frac{2}{3}})

Add a constant to the solution

= \frac{1}{320} (60 x^{2} (1 + 2 x)^{\frac{2}{3}} - 36 x (1 + 2 x)^{\frac{2}{3}} + 27 (1 + 2 x)^{\frac{2}{3}}) + C

\frac{\partial}{\partial y} (e^{x y} y^{2} + \frac{x}{y})

\frac{\partial}{\partial r} (r^{2} + s^{2})

\frac{d}{d x} (sin (x))

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

y = (7 - 3 x^{3})^{3}

What is the integral of (x^2)/((1+2x)^{1/3)} ?
The integral of (x^2)/((1+2x)^{1/3)} is 1/320 (60x^2(1+2x)^{2/3}-36x(1+2x)^{2/3}+27(1+2x)^{2/3})+C