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

The integral of \sqrt[3]{x}(x+x^2)' is 3/4 x^{4/3}(x+x^2)'-9/14 x^{7/3}+C

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

\int 3 x (x + x^{2})^{'} d x

\frac{3}{4} x^{\frac{4}{3}} (x + x^{2})' - \frac{9}{14} x^{\frac{7}{3}} + C

Solution steps

\int 3 x (x + x^{2})^{'} d x

Apply Integration By Parts

: \frac{3}{4} x^{\frac{4}{3}} (x + x^{2})^{'} - \int \frac{3}{2} x^{\frac{4}{3}} d x

= \frac{3}{4} x^{\frac{4}{3}} (x + x^{2})^{^{'}} - \int \frac{3}{2} x^{\frac{4}{3}} d x

\int \frac{3}{2} x^{\frac{4}{3}} d x = \frac{9}{14} x^{\frac{7}{3}}

= \frac{3}{4} x^{\frac{4}{3}} (x + x^{2})^{^{'}} - \frac{9}{14} x^{\frac{7}{3}}

Add a constant to the solution

= \frac{3}{4} x^{\frac{4}{3}} (x + x^{2})^{^{'}} - \frac{9}{14} x^{\frac{7}{3}} + C

\frac{d y}{d x} = \frac{y}{x} + 5 x + 1

\frac{\partial}{\partial x} (4 y x)

y^{^{'}} = 5 x e^{y}, y (0) = 0

\int_{0}^{\infty} x e^{- 4 x} d x

\int \frac{1}{x ^{2} - 256} d x

What is the integral of \sqrt[3]{x}(x+x^2)' ?
The integral of \sqrt[3]{x}(x+x^2)' is 3/4 x^{4/3}(x+x^2)'-9/14 x^{7/3}+C