What is the integral from 2 to 3 of 7/x+x^{-1} ?

The integral from 2 to 3 of 7/x+x^{-1} is 8(ln(3)-ln(2))

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integral from 2 to 3 of 7/x+x^{-1}

\int_{2}^{3} \frac{7}{x} + x^{- 1} d x

8 (ln (3) - ln (2))

Decimal Notation

3.24372 \dots

Solution steps

\int_{2}^{3} \frac{7}{x} + x^{- 1} d x

Simplify

\frac{7}{x} + x^{- 1} : \frac{8}{x}

= \int_{2}^{3} \frac{8}{x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int_{2}^{3} \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= 8 [ln ∣ x ∣]_{2}^{3}

Compute the boundaries:

ln (3) - ln (2)

= 8 (ln (3) - ln (2))

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What is the integral from 2 to 3 of 7/x+x^{-1} ?
The integral from 2 to 3 of 7/x+x^{-1} is 8(ln(3)-ln(2))