Solve ∫ 7x-1+3x+12x wrt x

$$\int{ 7x-1+3x+12x }d x$$

Evaluate

$$11x^{2}-x+С$$

Solution Steps

Integrate the sum term by term.

$$\int 7x\mathrm{d}x+\int -1\mathrm{d}x+\int 3x\mathrm{d}x+\int 12x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$7\int x\mathrm{d}x+\int -1\mathrm{d}x+3\int x\mathrm{d}x+12\int x\mathrm{d}x$$

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 7 times \frac{x^{2}}{2}.

$$\frac{7x^{2}}{2}+\int -1\mathrm{d}x+3\int x\mathrm{d}x+12\int x\mathrm{d}x$$

Find the integral of -1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{7x^{2}}{2}-x+3\int x\mathrm{d}x+12\int x\mathrm{d}x$$

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 3 times \frac{x^{2}}{2}.

$$\frac{7x^{2}}{2}-x+\frac{3x^{2}}{2}+12\int x\mathrm{d}x$$

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 12 times \frac{x^{2}}{2}.

$$\frac{7x^{2}}{2}-x+\frac{3x^{2}}{2}+6x^{2}$$

Simplify.

$$11x^{2}-x$$

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.

$$11x^{2}-x+С$$

Differentiate w.r.t. x

$$22x-1$$