What is the integral from 1 to 2 of (2/x-e^{3x}) ?

The integral from 1 to 2 of (2/x-e^{3x}) is 2ln(2)-(e^6-e^3)/3

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integral from 1 to 2 of (2/x-e^{3x})

\int_{1}^{2} (\frac{2}{x} - e^{3 x}) d x

2 ln (2) - \frac{e ^{6} - e ^{3}}{3}

Decimal

- 126.39479 \dots

Solution steps

\int_{1}^{2} \frac{2}{x} - e^{3 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} \frac{2}{x} d x - \int_{1}^{2} e^{3 x} d x

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

\int_{1}^{2} e^{3 x} d x = \frac{e ^{6} - e ^{3}}{3}

= 2 ln (2) - \frac{e ^{6} - e ^{3}}{3}

What is the integral from 1 to 2 of (2/x-e^{3x}) ?
The integral from 1 to 2 of (2/x-e^{3x}) is 2ln(2)-(e^6-e^3)/3