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

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

Evaluate

$$3\ln(3)+\frac{26}{3}\approx 11.962503533$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Use \int \frac{1}{x}\mathrm{d}x=\ln(|x|) from the table of common integrals to obtain the result.

$$3\ln(|x|)+\int x^{2}\mathrm{d}x$$

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}.

$$3\ln(|x|)+\frac{x^{3}}{3}$$

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.

$$3\ln(|3|)+\frac{3^{3}}{3}-\left(3\ln(|1|)+\frac{1^{3}}{3}\right)$$

Simplify.

$$3\ln(3)+\frac{26}{3}$$