What is the integral from 3 to 8 of x/(x^2-1) ?

The integral from 3 to 8 of x/(x^2-1) is 1/2 (ln(63)-3ln(2))

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integral from 3 to 8 of x/(x^2-1)

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

\frac{1}{2} (ln (63) - 3 ln (2))

Solution steps

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

Apply u-substitution

: \int_{8}^{63} \frac{1}{2 u} d u

= \int_{8}^{63} \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{8}^{63} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{8}^{63}

Compute the boundaries:

ln (63) - 3 ln (2)

= \frac{1}{2} (ln (63) - 3 ln (2))

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What is the integral from 3 to 8 of x/(x^2-1) ?
The integral from 3 to 8 of x/(x^2-1) is 1/2 (ln(63)-3ln(2))