What is the integral of x/(\sqrt[3]{x+12)} ?

The integral of x/(\sqrt[3]{x+12)} is 3/5 (x+12)^{5/3}-18(x+12)^{2/3}+C

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integral of x/(\sqrt[3]{x+12)}

\frac{3}{5} (x + 12)^{\frac{5}{3}} - 18 (x + 12)^{\frac{2}{3}} + C

Solution steps

Apply u-substitution

Expand

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

= \frac{3}{5} u^{\frac{5}{3}} - 18 u^{\frac{2}{3}}

Substitute back

u = x + 12

= \frac{3}{5} (x + 12)^{\frac{5}{3}} - 18 (x + 12)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{5} (x + 12)^{\frac{5}{3}} - 18 (x + 12)^{\frac{2}{3}} + C

f (x) = 6 x sin (x)

\frac{d y}{d x} = x x - 9

y = 3 - x, y = - 3 - x, y = x - 1

x \to 1 lim (4 x^{2} - 7 x + 5)

\int e^{x} cos (5 x) d x

What is the integral of x/(\sqrt[3]{x+12)} ?
The integral of x/(\sqrt[3]{x+12)} is 3/5 (x+12)^{5/3}-18(x+12)^{2/3}+C