Solve ∫ (3sqrt{x}+2/sqrt{x}) wrt x

$$\int ( 3 \sqrt { x } + \frac { 2 } { \sqrt { x } } ) d x$$

Evaluate

$$2x^{\frac{3}{2}}+4\sqrt{x}+С$$

Solution Steps

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

$$3\int \sqrt{x}\mathrm{d}x+2\int \frac{1}{\sqrt{x}}\mathrm{d}x$$

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}.

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

Rewrite \frac{1}{\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{1}{2}}}{\frac{1}{2}}. Simplify and convert from exponential to radical form. Multiply 2 times 2\sqrt{x}.

$$2x^{\frac{3}{2}}+4\sqrt{x}$$

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.

$$2x^{\frac{3}{2}}+4\sqrt{x}+С$$

Differentiate w.r.t. x

$$\frac{3x+2}{\sqrt{x}}$$