Solve ∫ (1-sqrt[3]{x^2})^2 wrt x

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

Evaluate

$$\frac{3x^{\frac{7}{3}}}{7}-\frac{6x^{\frac{5}{3}}}{5}+x+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-\sqrt[3]{x^{2}}\right)^{2}.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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 -2 times \frac{3x^{\frac{5}{3}}}{5}.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{4}{3}}\mathrm{d}x with \frac{3x^{\frac{7}{3}}}{7}.

$$x-\frac{6x^{\frac{5}{3}}}{5}+\frac{3x^{\frac{7}{3}}}{7}$$

Simplify.

$$\frac{3x^{\frac{7}{3}}}{7}-\frac{6x^{\frac{5}{3}}}{5}+x$$

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.

$$\frac{3x^{\frac{7}{3}}}{7}-\frac{6x^{\frac{5}{3}}}{5}+x+С$$

Differentiate w.r.t. x

$$\left(x^{\frac{2}{3}}-1\right)^{2}$$