What is the integral from 3 to 7 of x/(x^2-2) ?

The integral from 3 to 7 of x/(x^2-2) is 1/2 (ln(47)-ln(7))

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integral from 3 to 7 of x/(x^2-2)

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

\frac{1}{2} (ln (47) - ln (7))

Decimal Notation

0.95211 \dots

Solution steps

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

Apply u-substitution

: \int_{7}^{47} \frac{1}{2 u} d u

= \int_{7}^{47} \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_{7}^{47} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{7}^{47}

Compute the boundaries:

ln (47) - ln (7)

= \frac{1}{2} (ln (47) - ln (7))

y = x^{2} - 5 x + 5 (0.5)

F (s) = ln (ln (6 s))

\int_{0}^{\frac{π}{2}} sec (x) d x

t \to 0 lim (\frac{tan ( 10 t )}{sin ( 2 t )})

\frac{d}{d x} (x \cdot e^{3 x})

What is the integral from 3 to 7 of x/(x^2-2) ?
The integral from 3 to 7 of x/(x^2-2) is 1/2 (ln(47)-ln(7))