What is the integral of x/(\sqrt[3]{x+4)} ?

The integral of x/(\sqrt[3]{x+4)} is 3/5 (x+4)^{5/3}-6(x+4)^{2/3}+C

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integral of x/(\sqrt[3]{x+4)}

\int \frac{x}{3 x + 4} d x

\frac{3}{5} (x + 4)^{\frac{5}{3}} - 6 (x + 4)^{\frac{2}{3}} + C

Solution steps

\int \frac{x}{3 x + 4} d x

Apply u-substitution

: \int \frac{u - 4}{3 u} d u

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

Expand

\frac{u - 4}{3 u} : u^{\frac{2}{3}} - \frac{4}{3 u}

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

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

\int \frac{4}{3 u} d u = 6 u^{\frac{2}{3}}

= \frac{3}{5} u^{\frac{5}{3}} - 6 u^{\frac{2}{3}}

Substitute back

u = x + 4

= \frac{3}{5} (x + 4)^{\frac{5}{3}} - 6 (x + 4)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{5} (x + 4)^{\frac{5}{3}} - 6 (x + 4)^{\frac{2}{3}} + C

x \to 3 lim (\frac{1}{( x - 3 ) ^{8}})

x sin (\frac{3}{x})

x \to - 3 lim (\frac{x ^{2} - 9}{3 + x})

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

\int_{3}^{5} \frac{x}{x ^{2} - 3} d x

What is the integral of x/(\sqrt[3]{x+4)} ?
The integral of x/(\sqrt[3]{x+4)} is 3/5 (x+4)^{5/3}-6(x+4)^{2/3}+C