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

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

Evaluate

$$e^{2}+\ln(2)-e\approx 5.363921451$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Use \int e^{x}\mathrm{d}x=e^{x} from the table of common integrals to obtain the result.

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

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

$$e^{x}+\ln(|x|)$$

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.

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

Simplify.

$$e^{2}+\ln(2)-e$$