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

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

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

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

Solution steps

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

\int 1 d x = x

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

Add a constant to the solution

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

\frac{e ^{- 2 s}}{s - 1}

\frac{d y}{d t} = \frac{2}{t} y + t^{3} e^{t}

\int 3^{- 6 x} d x

\frac{3}{s ( s ^{2} + 2 s + 4 )}

\int y^{- 5} d y

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