Solve ∫ (from 1 to 4) of (2+1/x^2) wrt x

$$\int _ { 1 } ^ { 4 } ( 2 + \frac { 1 } { x ^ { 2 } } ) d x$$

Evaluate

$$\frac{27}{4}=6.75$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

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

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

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}.

$$2x-\frac{1}{x}$$

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.

$$2\times 4-4^{-1}-\left(2\times 1-1^{-1}\right)$$

Simplify.

$$\frac{27}{4}$$