Solve ∫ 8(sqrt[3]{x}-sqrt[3]{x^2}+2x) wrt x

$$\int 8 ( \sqrt[ 3 ] { x } - \sqrt[ 3 ] { x ^ { 2 } } + 2 x ) d x$$

Evaluate

$$8x^{2}-\frac{24x^{\frac{5}{3}}}{5}+6x^{\frac{4}{3}}+С$$

Solution Steps

Use the distributive property to multiply 8 by \sqrt[3]{x}-\sqrt[3]{x^{2}}+2x.

$$\int 8\left(\sqrt[3]{x}-\sqrt[3]{x^{2}}\right)+16x\mathrm{d}x$$

Use the distributive property to multiply 8 by \sqrt[3]{x}-\sqrt[3]{x^{2}}.

$$\int 8\sqrt[3]{x}-8\sqrt[3]{x^{2}}+16x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 8\sqrt[3]{x}\mathrm{d}x+\int -8x^{\frac{2}{3}}\mathrm{d}x+\int 16x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$8\int \sqrt[3]{x}\mathrm{d}x-8\int x^{\frac{2}{3}}\mathrm{d}x+16\int x\mathrm{d}x$$

Rewrite \sqrt[3]{x} as x^{\frac{1}{3}}. Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{1}{3}}\mathrm{d}x with \frac{x^{\frac{4}{3}}}{\frac{4}{3}}. Simplify. Multiply 8 times \frac{3x^{\frac{4}{3}}}{4}.

$$6x^{\frac{4}{3}}-8\int x^{\frac{2}{3}}\mathrm{d}x+16\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{2}{3}}\mathrm{d}x with \frac{3x^{\frac{5}{3}}}{5}. Multiply -8 times \frac{3x^{\frac{5}{3}}}{5}.

$$6x^{\frac{4}{3}}-\frac{24x^{\frac{5}{3}}}{5}+16\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply 16 times \frac{x^{2}}{2}.

$$6x^{\frac{4}{3}}-\frac{24x^{\frac{5}{3}}}{5}+8x^{2}$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$6x^{\frac{4}{3}}-\frac{24x^{\frac{5}{3}}}{5}+8x^{2}+С$$

Differentiate w.r.t. x

$$16x-8x^{\frac{2}{3}}+8\sqrt[3]{x}$$