Solve ∫ ((8x^2)+(2x^3)+(86)) wrt x=?

$$\int ( ( 8 x ^ { 2 } ) + ( 2 x ^ { 3 } ) + ( 86 ) ) d x = ?$$

Evaluate

$$\frac{x^{4}}{2}+\frac{8x^{3}}{3}+86x+С$$

Solution Steps

Integrate the sum term by term.

$$\int 8x^{2}\mathrm{d}x+\int 2x^{3}\mathrm{d}x+\int 86\mathrm{d}x$$

Factor out the constant in each of the terms.

$$8\int x^{2}\mathrm{d}x+2\int x^{3}\mathrm{d}x+\int 86\mathrm{d}x$$

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}.

$$\frac{8x^{3}}{3}+2\int x^{3}\mathrm{d}x+\int 86\mathrm{d}x$$

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}. Multiply 2 times \frac{x^{4}}{4}.

$$\frac{8x^{3}}{3}+\frac{x^{4}}{2}+\int 86\mathrm{d}x$$

Find the integral of 86 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{8x^{3}}{3}+\frac{x^{4}}{2}+86x$$

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.

$$\frac{8x^{3}}{3}+\frac{x^{4}}{2}+86x+С$$

Differentiate w.r.t. x

$$2\left(x^{3}+4x^{2}+43\right)$$