Solve ∫ (from 1 to 4) of left(-x+4/xright) wrt x

$$\int_{ 1 }^{ 4 } \left( -x+ \frac{ 4 }{ x } \right) d x$$

Evaluate

$$8\ln(2)-\frac{15}{2}\approx -1.954822556$$

Solution Steps

Evaluate the indefinite integral first.

$$\int -x+\frac{4}{x}\mathrm{d}x$$

Integrate the sum term by term.

$$\int -x\mathrm{d}x+\int \frac{4}{x}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-\int x\mathrm{d}x+4\int \frac{1}{x}\mathrm{d}x$$

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

$$-\frac{x^{2}}{2}+4\int \frac{1}{x}\mathrm{d}x$$

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

$$-\frac{x^{2}}{2}+4\ln(|x|)$$

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.

$$-\frac{4^{2}}{2}+4\ln(|4|)-\left(-\frac{1^{2}}{2}+4\ln(|1|)\right)$$

Simplify.

$$-\frac{15}{2}+8\ln(2)$$