Solve ∫ (6x^2-3sqrt{x}) wrt x

$$\int ( 6 x ^ { 2 } - 3 \sqrt { x } ) d x$$

Evaluate

$$2x^{3}-2x^{\frac{3}{2}}+С$$

Solution Steps

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

$$6\int x^{2}\mathrm{d}x-3\int \sqrt{x}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 6 times \frac{x^{3}}{3}.

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

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}.

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

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.

$$2x^{3}-2x^{\frac{3}{2}}+С$$

Differentiate w.r.t. x

$$6x^{2}-3\sqrt{x}$$