Solve ∫ (from 3 to 4) of [x^2+1/x^2] wrt x

$$\int _ { 3 } ^ { 4 } [ x ^ { 2 } + \frac { 1 } { x ^ { 2 } } ] d x$$

Evaluate

$$\frac{149}{12}\approx 12.416666667$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

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

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

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}.

$$\frac{x^{3}}{3}-\frac{1}{x}$$

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.

$$\frac{4^{3}}{3}-4^{-1}-\left(\frac{3^{3}}{3}-3^{-1}\right)$$

Simplify.

$$\frac{149}{12}$$