Solve ∫ (from 2 to 3) of 8/x^-2+9+20x wrt x

$$\int_{ 2 }^{ 3 } \frac{ 8 }{ { x }^{ -2 } } +9+20x d x$$

Evaluate

$$\frac{329}{3}\approx 109.666666667$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 9+20x times \frac{x^{-2}}{x^{-2}}.

$$\int _{2}^{3}\frac{8}{x^{-2}}+\frac{\left(9+20x\right)x^{-2}}{x^{-2}}\mathrm{d}x$$

Since \frac{8}{x^{-2}} and \frac{\left(9+20x\right)x^{-2}}{x^{-2}} have the same denominator, add them by adding their numerators.

$$\int _{2}^{3}\frac{8+\left(9+20x\right)x^{-2}}{x^{-2}}\mathrm{d}x$$

Do the multiplications in 8+\left(9+20x\right)x^{-2}.

$$\int _{2}^{3}\frac{8+9x^{-2}+20\times \frac{1}{x}}{x^{-2}}\mathrm{d}x$$

Combine like terms in 8+9x^{-2}+20\times \frac{1}{x}.

$$\int _{2}^{3}\frac{8+\left(9+20x\right)x^{-2}}{x^{-2}}\mathrm{d}x$$

Factor the expressions that are not already factored in \frac{8+\left(9+20x\right)x^{-2}}{x^{-2}}.

$$\int _{2}^{3}\frac{8x^{-2}\left(x-\left(-\frac{1}{4}\sqrt{7}-\frac{5}{4}\right)\right)\left(x-\left(\frac{1}{4}\sqrt{7}-\frac{5}{4}\right)\right)}{x^{-2}}\mathrm{d}x$$

Cancel out x^{-2} in both numerator and denominator.

$$\int _{2}^{3}8\left(x-\left(-\frac{1}{4}\sqrt{7}-\frac{5}{4}\right)\right)\left(x-\left(\frac{1}{4}\sqrt{7}-\frac{5}{4}\right)\right)\mathrm{d}x$$

Expand the expression.

$$\int _{2}^{3}8x^{2}+20x+9\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 8x^{2}+20x+9\mathrm{d}x$$

Integrate the sum term by term.

$$\int 8x^{2}\mathrm{d}x+\int 20x\mathrm{d}x+\int 9\mathrm{d}x$$

Factor out the constant in each of the terms.

$$8\int x^{2}\mathrm{d}x+20\int x\mathrm{d}x+\int 9\mathrm{d}x$$

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}.

$$\frac{8x^{3}}{3}+20\int x\mathrm{d}x+\int 9\mathrm{d}x$$

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

$$\frac{8x^{3}}{3}+10x^{2}+\int 9\mathrm{d}x$$

Find the integral of 9 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{8x^{3}}{3}+10x^{2}+9x$$

Simplify.

$$\frac{8x^{3}}{3}+9x+10x^{2}$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$\frac{8}{3}\times 3^{3}+9\times 3+10\times 3^{2}-\left(\frac{8}{3}\times 2^{3}+9\times 2+10\times 2^{2}\right)$$

Simplify.

$$\frac{329}{3}$$