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

The integral of x\sqrt[3]{x-1} is 3/7 (x-1)^{7/3}+3/4 (x-1)^{4/3}+C

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

\frac{3}{7} (x - 1)^{\frac{7}{3}} + \frac{3}{4} (x - 1)^{\frac{4}{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{4}{3}} d u = \frac{3}{7} u^{\frac{7}{3}}

= \frac{3}{7} u^{\frac{7}{3}} + \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = x - 1

= \frac{3}{7} (x - 1)^{\frac{7}{3}} + \frac{3}{4} (x - 1)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{7} (x - 1)^{\frac{7}{3}} + \frac{3}{4} (x - 1)^{\frac{4}{3}} + C

e^{\frac{x}{2}}

3 x^{2} - 4

\frac{d y}{d x}, y^{3} = 4 x

\int (y ln (y)) d y

\frac{\partial}{\partial x} ((x y - 3) e^{- x y})

What is the integral of x\sqrt[3]{x-1} ?
The integral of x\sqrt[3]{x-1} is 3/7 (x-1)^{7/3}+3/4 (x-1)^{4/3}+C