What is the integral from 2 to 6 of (e^{-2x}+ln(x)) ?

The integral from 2 to 6 of (e^{-2x}+ln(x)) is (e^8-1)/(2e^{12)}+2ln(108)-4

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integral from 2 to 6 of (e^{-2x}+ln(x))

\int_{2}^{6} (e^{- 2 x} + ln (x)) d x

\frac{e ^{8} - 1}{2 e ^{12}} + 2 ln (108) - 4

Decimal Notation

5.37341 \dots

Solution steps

\int_{2}^{6} e^{- 2 x} + ln (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_{2}^{6} e^{- 2 x} d x + \int_{2}^{6} ln (x) d x

\int_{2}^{6} e^{- 2 x} d x = \frac{e ^{8} - 1}{2 e ^{12}}

\int_{2}^{6} ln (x) d x = 2 ln (108) - 4

= \frac{e ^{8} - 1}{2 e ^{12}} + 2 ln (108) - 4

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What is the integral from 2 to 6 of (e^{-2x}+ln(x)) ?
The integral from 2 to 6 of (e^{-2x}+ln(x)) is (e^8-1)/(2e^{12)}+2ln(108)-4