What is the integral of 1/(\sqrt[3]{x-3)} ?

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

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integral of 1/(\sqrt[3]{x-3)}

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

Solution steps

Apply u-substitution

Apply the Power Rule

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

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

Substitute back

u = x - 3

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

Add a constant to the solution

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

x^{2} + 25

\frac{d}{d y} (e^{x^{2} - y^{2}} cos (2 x y))

x \to 0 lim (\frac{5 x ^{2} - 2 x}{x})

\int \frac{250}{( x ^{2} + 25 ) ^{2}} d x

y = e^{s i n (x)}

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