What is the integral from-5 to-2 of (7e^x+2/x) ?

The integral from-5 to-2 of (7e^x+2/x) is 7(1/(e^2)-1/(e^5))+2(ln(2)-ln(5))

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integral from-5 to-2 of (7e^x+2/x)

\int_{- 5}^{- 2} (7 e^{x} + \frac{2}{x}) d x

7 (\frac{1}{e ^{2}} - \frac{1}{e ^{5}}) + 2 (ln (2) - ln (5))

Decimal Notation

- 0.93240 \dots

Solution steps

\int_{- 5}^{- 2} 7 e^{x} + \frac{2}{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_{- 5}^{- 2} 7 e^{x} d x + \int_{- 5}^{- 2} \frac{2}{x} d x

\int_{- 5}^{- 2} 7 e^{x} d x = 7 (\frac{1}{e ^{2}} - \frac{1}{e ^{5}})

\int_{- 5}^{- 2} \frac{2}{x} d x = 2 (ln (2) - ln (5))

= 7 (\frac{1}{e ^{2}} - \frac{1}{e ^{5}}) + 2 (ln (2) - ln (5))

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What is the integral from-5 to-2 of (7e^x+2/x) ?
The integral from-5 to-2 of (7e^x+2/x) is 7(1/(e^2)-1/(e^5))+2(ln(2)-ln(5))