Solve f ( x ) = ∫ (from ln x to 2) of (e^t+t)dt

$$f ( x ) = \int _ { \ln x } ^ { 2 } ( e ^ { t } + t ) d t$$

Evaluate

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

Solution Steps

Evaluate the indefinite integral first.

$$\int e^{t}+t\mathrm{d}t$$

Integrate the sum term by term.

$$\int e^{t}\mathrm{d}t+\int t\mathrm{d}t$$

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

$$e^{t}+\int t\mathrm{d}t$$

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

$$e^{t}+\frac{t^{2}}{2}$$

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.

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

Simplify.

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

Differentiate w.r.t. x

$$-\frac{\ln(x)}{x}-1$$