Solve ∫ 6 wrt x/sqrt[3]{x}

$$\int \frac { 6 d x } { \sqrt[ 3 ] { x } }$$

Evaluate

$$9x^{\frac{2}{3}}+С$$

Solution Steps

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

$$6\int \frac{1}{\sqrt[3]{x}}\mathrm{d}x$$

Rewrite \frac{1}{\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{2}{3}}}{\frac{2}{3}}. Simplify. Multiply 6 times \frac{3x^{\frac{2}{3}}}{2}.

$$9x^{\frac{2}{3}}$$

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.

$$9x^{\frac{2}{3}}+С$$

Differentiate w.r.t. x

$$\frac{6}{\sqrt[3]{x}}$$