What is the integral from 1 to 2 of 3^x+1/x ?

The integral from 1 to 2 of 3^x+1/x is 6/(ln(3))+ln(2)

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integral from 1 to 2 of 3^x+1/x

\int_{1}^{2} 3^{x} + \frac{1}{x} d x

\frac{6}{ln ( 3 )} + ln (2)

Decimal Notation

6.15458 \dots

Solution steps

\int_{1}^{2} 3^{x} + \frac{1}{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} 3^{x} d x + \int_{1}^{2} \frac{1}{x} d x

\int_{1}^{2} 3^{x} d x = \frac{6}{ln ( 3 )}

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

= \frac{6}{ln ( 3 )} + ln (2)

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What is the integral from 1 to 2 of 3^x+1/x ?
The integral from 1 to 2 of 3^x+1/x is 6/(ln(3))+ln(2)