Solve ∫ (from 1 to 2) of (3/x^2+x^2+2) wrt x

$$\int_{ 1 }^{ 2 } ( \frac{ 3 }{ { x }^{ 2 } } + { x }^{ 2 } +2) d x$$

Evaluate

$$\frac{35}{6}\approx 5.833333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int \frac{1}{x^{2}}\mathrm{d}x with -\frac{1}{x}. Multiply 3 times -\frac{1}{x}.

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

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}.

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

Find the integral of 2 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

$$-3\times 2^{-1}+\frac{2^{3}}{3}+2\times 2-\left(-3\times 1^{-1}+\frac{1^{3}}{3}+2\times 1\right)$$

Simplify.

$$\frac{35}{6}$$