Solve ∫ (from 1 to 6) of (1/x^2-1) wrt x

$$\int _ { 1 } ^ { 6 } ( \frac { 1 } { x ^ { 2 } } - 1 ) d x$$

Evaluate

$$-\frac{25}{6}\approx -4.166666667$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

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}.

$$-\frac{1}{x}+\int -1\mathrm{d}x$$

Find the integral of -1 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

$$-6^{-1}-6-\left(-1^{-1}-1\right)$$

Simplify.

$$-\frac{25}{6}$$