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

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

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

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

= \frac{3}{2} (x - 1)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{2} (x - 1)^{\frac{2}{3}} + C

\int (y e^{2 x y} + x) d x

\frac{d y}{d x} = (y + 3)

y = x^{3} - 5 x + 1, (3, 13)

f^{^{''}} (x) = \frac{4}{x ^{2}}

\int \frac{1}{x ^{2} 49 x ^{2} + 16} d x

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