What is the integral of x/(\sqrt[3]{x+2)} ?

The integral of x/(\sqrt[3]{x+2)} is 3/5 (x+2)^{5/3}-3(x+2)^{2/3}+C

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integral of x/(\sqrt[3]{x+2)}

\int \frac{x}{3 x + 2} d x

\frac{3}{5} (x + 2)^{\frac{5}{3}} - 3 (x + 2)^{\frac{2}{3}} + C

Solution steps

\int \frac{x}{3 x + 2} d x

Apply u-substitution

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

Expand

\frac{u - 2}{3 u} : u^{\frac{2}{3}} - \frac{2}{3 u}

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

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

\int u^{\frac{2}{3}} d u = \frac{3}{5} u^{\frac{5}{3}}

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

= \frac{3}{5} u^{\frac{5}{3}} - 3 u^{\frac{2}{3}}

Substitute back

u = x + 2

= \frac{3}{5} (x + 2)^{\frac{5}{3}} - 3 (x + 2)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{5} (x + 2)^{\frac{5}{3}} - 3 (x + 2)^{\frac{2}{3}} + C

\frac{\partial}{\partial x} (x^{3} + 4 x^{2} y + 2 y^{3})

t an g e n t f (x) = \frac{2}{x - 1}, (3, 1)

\frac{d}{d y} (\frac{1}{3} y (y - 3))

\frac{d}{d x} (x^{l n (x^{2} + 1)})

\frac{d ^{2}}{d x ^{2}} (e^{s i n (x)})

What is the integral of x/(\sqrt[3]{x+2)} ?
The integral of x/(\sqrt[3]{x+2)} is 3/5 (x+2)^{5/3}-3(x+2)^{2/3}+C