What is the integral from 1 to 2e of (2/x+e^x) ?

The integral from 1 to 2e of (2/x+e^x) is 2ln(2e)+e^{2e}-e

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integral from 1 to 2e of (2/x+e^x)

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

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

Decimal

230.31967 \dots

Solution steps

\int_{1}^{2 e} \frac{2}{x} + e^{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2 e} \frac{2}{x} d x + \int_{1}^{2 e} e^{x} d x

\int_{1}^{2 e} \frac{2}{x} d x = 2 ln (2 e)

\int_{1}^{2 e} e^{x} d x = e^{2 e} - e

= 2 ln (2 e) + e^{2 e} - e

What is the integral from 1 to 2e of (2/x+e^x) ?
The integral from 1 to 2e of (2/x+e^x) is 2ln(2e)+e^{2e}-e