Solve ∫ (1-x+x^3-1/sqrt[3]{x^2}) wrt x

$$\int ( 1 - x + x ^ { 3 } - \frac { 1 } { \sqrt[ 3 ] { x ^ { 2 } } } ) d x$$

Evaluate

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

Solution Steps

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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 -1 times \frac{x^{2}}{2}.

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

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

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

Rewrite \frac{1}{x^{\frac{2}{3}}} as x^{-\frac{2}{3}}. 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{x^{\frac{1}{3}}}{\frac{1}{3}}. Simplify and convert from exponential to radical form. Multiply -1 times 3\sqrt[3]{x}.

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

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.

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

Differentiate w.r.t. x

$$x^{3}-x+1-\frac{1}{x^{\frac{2}{3}}}$$