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

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

Evaluate

$$-\frac{6x^{\frac{5}{3}}}{25}+2x^{\frac{3}{2}}+\sqrt[5]{125}x+С$$

Solution Steps

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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

$$2x^{\frac{3}{2}}-\frac{6x^{\frac{5}{3}}}{25}+\int 5^{\frac{3}{5}}\mathrm{d}x$$

Find the integral of 5^{\frac{3}{5}} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$2x^{\frac{3}{2}}-\frac{6x^{\frac{5}{3}}}{25}+5^{\frac{3}{5}}x$$

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.

$$2x^{\frac{3}{2}}-\frac{6x^{\frac{5}{3}}}{25}+5^{\frac{3}{5}}x+С$$

Differentiate w.r.t. x

$$-\frac{2x^{\frac{2}{3}}}{5}+3\sqrt{x}+\sqrt[5]{125}$$