Solve ∫ (4x^2-2sqrt[3]{x}+8/x^3) wrt x

$$\int{ (4 { x }^{ 2 } -2 \sqrt[3]{ x } + \frac{ 8 }{ { x }^{ 3 } } ) }d x$$

Evaluate

$$\frac{4x^{3}}{3}-\frac{3x^{\frac{4}{3}}}{2}-\frac{4}{x^{2}}+С$$

Solution Steps

Integrate the sum term by term.

$$\int 4x^{2}\mathrm{d}x+\int -2\sqrt[3]{x}\mathrm{d}x+\int \frac{8}{x^{3}}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x^{2}\mathrm{d}x-2\int \sqrt[3]{x}\mathrm{d}x+8\int \frac{1}{x^{3}}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 4 times \frac{x^{3}}{3}.

$$\frac{4x^{3}}{3}-2\int \sqrt[3]{x}\mathrm{d}x+8\int \frac{1}{x^{3}}\mathrm{d}x$$

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 -2 times \frac{3x^{\frac{4}{3}}}{4}.

$$\frac{4x^{3}}{3}-\frac{3x^{\frac{4}{3}}}{2}+8\int \frac{1}{x^{3}}\mathrm{d}x$$

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

$$\frac{4x^{3}}{3}-\frac{3x^{\frac{4}{3}}}{2}-\frac{4}{x^{2}}$$

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.

$$\frac{4x^{3}}{3}-\frac{3x^{\frac{4}{3}}}{2}-\frac{4}{x^{2}}+С$$

Differentiate w.r.t. x

$$4x^{2}-2\sqrt[3]{x}+\frac{8}{x^{3}}$$