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

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

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

Solution steps

Apply u-substitution

: \int \frac{3 u ^{4}}{u ^{2} + 1} d u

= \int \frac{3 u ^{4}}{u ^{2} + 1} d u

Take the constant out:

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

= 3 \cdot \int \frac{u ^{4}}{u ^{2} + 1} d u

Long division

\frac{u ^{4}}{u ^{2} + 1} : u^{2} - 1 + \frac{1}{u ^{2} + 1}

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

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

\int 1 d u = u

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 3 (\frac{u ^{3}}{3} - u + arctan (u))

Substitute back

Simplify

Add a constant to the solution

\frac{d}{d x} ((\frac{1}{x ^{2}} - \frac{9}{x ^{4}}) (x + 5 x^{3}))

x \to 0 lim (2 e^{3 x})

f (x) = \frac{12 x}{x ^{2} - 4}, (4, 4)

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

\frac{8 x ^{2} + 9}{x ^{2} + 5}

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