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

\int sin (7 x) d x

y^{^{''}} + 4 y = 4 t

f (x) = (7^{- 3 x})

\int (4 x^{2} - \frac{2}{x}) d x

f (x) = - x^{2} + 1

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