Solve ∫ 2sqrt[3]{x^2}+4/x^4+3/x^5 wrt x

$$\int{ 2 \sqrt[ 3 ]{ { x }^{ 2 } } + \frac{ 4 }{ { x }^{ 4 } } + \frac{ 3 }{ { x }^{ 5 } } }d x$$

Evaluate

$$\frac{6x^{\frac{5}{3}}}{5}-\frac{4}{3x^{3}}-\frac{3}{4x^{4}}+С$$

Solution Steps

Integrate the sum term by term.

$$\int 2x^{\frac{2}{3}}\mathrm{d}x+\int \frac{4}{x^{4}}\mathrm{d}x+\int \frac{3}{x^{5}}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{\frac{2}{3}}\mathrm{d}x+4\int \frac{1}{x^{4}}\mathrm{d}x+3\int \frac{1}{x^{5}}\mathrm{d}x$$

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

$$\frac{6x^{\frac{5}{3}}}{5}+4\int \frac{1}{x^{4}}\mathrm{d}x+3\int \frac{1}{x^{5}}\mathrm{d}x$$

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

$$\frac{6x^{\frac{5}{3}}}{5}-\frac{\frac{4}{x^{3}}}{3}+3\int \frac{1}{x^{5}}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int \frac{1}{x^{5}}\mathrm{d}x with -\frac{1}{4x^{4}}. Multiply 3 times -\frac{1}{4x^{4}}.

$$\frac{6x^{\frac{5}{3}}}{5}-\frac{\frac{4}{x^{3}}}{3}-\frac{\frac{3}{x^{4}}}{4}$$

Simplify.

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

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.

$$\frac{6x^{\frac{5}{3}}}{5}+\frac{-\frac{3}{4}-\frac{4x}{3}}{x^{4}}+С$$

Differentiate w.r.t. x

$$\frac{2x^{\frac{17}{3}}+4x+3}{x^{5}}$$