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

The integral of (x^2+1)/(\sqrt[3]{x)} is 3/8 x^{8/3}+3/2 x^{2/3}+C

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

\frac{3}{8} x^{\frac{8}{3}} + \frac{3}{2} x^{\frac{2}{3}} + C

Solution steps

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 x^{\frac{5}{3}} d x = \frac{3}{8} x^{\frac{8}{3}}

= \frac{3}{8} x^{\frac{8}{3}} + \frac{3}{2} x^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{8} x^{\frac{8}{3}} + \frac{3}{2} x^{\frac{2}{3}} + C

\frac{\partial}{\partial y} ((A^{2} x^{2} + B^{2} y^{2})^{4})

\int \frac{e ^{\frac{4}{x}}}{x ^{2}} d x

x \to - \infty lim (xx)

x \frac{d y}{d x} - (1 + x) y = y^{2}

What is the integral of (x^2+1)/(\sqrt[3]{x)} ?
The integral of (x^2+1)/(\sqrt[3]{x)} is 3/8 x^{8/3}+3/2 x^{2/3}+C