Solve ∫ sqrt[3]{32x^4 wrt x

$$\int \sqrt[ 3 ] { 32 x ^ { 4 } } d x$$

Evaluate

$$\frac{6\sqrt[3]{4}x^{\frac{7}{3}}}{7}+С$$

Solution Steps

Factor out the constant using \int af\left(x\right)\mathrm{d}x=a\int f\left(x\right)\mathrm{d}x.

$$\sqrt[3]{32}\int \sqrt[3]{x^{4}}\mathrm{d}x$$

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}.

$$\sqrt[3]{32}\times \frac{3x^{\frac{7}{3}}}{7}$$

Simplify.

$$\frac{6\sqrt[3]{4}x^{\frac{7}{3}}}{7}$$

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.

$$\frac{6\sqrt[3]{4}x^{\frac{7}{3}}}{7}+С$$

Differentiate w.r.t. x

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