Solve ∫ 8x^2+6x+3-int3x

$$\int 8 x ^ { 2 } + 6 x + 3 - \int 3 x$$

Evaluate

$$Сx+\frac{13x^{3}}{6}+3x^{2}+3x+С_{1}$$

Solution Steps

Integrate the sum term by term.

$$\int 8x^{2}\mathrm{d}x+\int 6x\mathrm{d}x+\int 3\mathrm{d}x+\int -\int 3x\mathrm{d}x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$8\int x^{2}\mathrm{d}x+6\int x\mathrm{d}x+\int 3\mathrm{d}x-\int \int 3x\mathrm{d}x\mathrm{d}x$$

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

$$\frac{8x^{3}}{3}+6\int x\mathrm{d}x+\int 3\mathrm{d}x-\int \int 3x\mathrm{d}x\mathrm{d}x$$

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

$$\frac{8x^{3}}{3}+3x^{2}+\int 3\mathrm{d}x-\int \int 3x\mathrm{d}x\mathrm{d}x$$

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

$$\frac{8x^{3}}{3}+3x^{2}+3x-\int \int 3x\mathrm{d}x\mathrm{d}x$$

Simplify.

$$\frac{8x^{3}}{3}+3x^{2}+3x-\left(Сx+\frac{x^{3}}{2}\right)$$

Simplify.

$$Сx+\frac{13x^{3}}{6}+3x^{2}+3x$$

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.

$$Сx+\frac{13x^{3}}{6}+3x^{2}+3x+С$$

Differentiate w.r.t. x

$$\frac{13x^{2}}{2}+6x+С$$