Solve ∫ (x^3-2x+sqrt{x}-12) wrt x=

$$\int ( x ^ { 3 } - 2 x + \sqrt { x } - 12 ) d x =$$

Evaluate

$$\frac{x^{4}}{4}-x^{2}+\frac{2x^{\frac{3}{2}}}{3}-12x+С$$

Solution Steps

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int -2x\mathrm{d}x+\int \sqrt{x}\mathrm{d}x+\int -12\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x-2\int x\mathrm{d}x+\int \sqrt{x}\mathrm{d}x+\int -12\mathrm{d}x$$

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

$$\frac{x^{4}}{4}-2\int x\mathrm{d}x+\int \sqrt{x}\mathrm{d}x+\int -12\mathrm{d}x$$

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

$$\frac{x^{4}}{4}-x^{2}+\int \sqrt{x}\mathrm{d}x+\int -12\mathrm{d}x$$

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.

$$\frac{x^{4}}{4}-x^{2}+\frac{2x^{\frac{3}{2}}}{3}+\int -12\mathrm{d}x$$

Find the integral of -12 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{4}}{4}-x^{2}+\frac{2x^{\frac{3}{2}}}{3}-12x$$

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.

$$\frac{x^{4}}{4}-x^{2}+\frac{2x^{\frac{3}{2}}}{3}-12x+С$$

Differentiate w.r.t. x

$$x^{3}-2x+\sqrt{x}-12$$