What is the integral of 1/(\sqrt[4]{x+2)} ?

The integral of 1/(\sqrt[4]{x+2)} is 4/3 (x+2)^{3/4}+C

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integral of 1/(\sqrt[4]{x+2)}

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

Solution steps

Apply u-substitution

Apply the Power Rule

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

= \frac{4}{3} u^{\frac{3}{4}}

Substitute back

u = x + 2

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

Add a constant to the solution

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

\frac{d}{d x} (x + \frac{64}{x})

\int \frac{3}{x ^{2} ( x ^{2} + 25 )} d x

\int \frac{1}{36 + x ^{2}} d x

\frac{27}{x ^{2} + 9}

\frac{\partial}{\partial x} (- 2 e^{4 y - x^{2} - y^{2}} x)

What is the integral of 1/(\sqrt[4]{x+2)} ?
The integral of 1/(\sqrt[4]{x+2)} is 4/3 (x+2)^{3/4}+C