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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int_{0}^{8} \frac{1}{3 x} d x

\int_{2}^{6} (x + 1) d x

\int_{1}^{x} (t^{2}) d t

\int_{0}^{4} x (4 - x) d x

\int_{0}^{20} ∣ x - 10 ∣ d x

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