Solve ∫ (6/x^3-2sqrt[5]{x^3}) wrt x

$$\int ( \frac { 6 } { x ^ { 3 } } - 2 \sqrt[ 5 ] { x ^ { 3 } } ) d x$$

Evaluate

$$-\frac{5x^{\frac{8}{5}}}{4}-\frac{3}{x^{2}}+С$$

Solution Steps

Integrate the sum term by term.

$$\int \frac{6}{x^{3}}\mathrm{d}x+\int -2x^{\frac{3}{5}}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$6\int \frac{1}{x^{3}}\mathrm{d}x-2\int x^{\frac{3}{5}}\mathrm{d}x$$

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

$$-\frac{3}{x^{2}}-2\int x^{\frac{3}{5}}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{3}{5}}\mathrm{d}x with \frac{5x^{\frac{8}{5}}}{8}. Multiply -2 times \frac{5x^{\frac{8}{5}}}{8}.

$$-\frac{3}{x^{2}}-\frac{5x^{\frac{8}{5}}}{4}$$

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.

$$-\frac{3}{x^{2}}-\frac{5x^{\frac{8}{5}}}{4}+С$$

Differentiate w.r.t. x

$$-2x^{\frac{3}{5}}+\frac{6}{x^{3}}$$