What is the integral from 4 to 7 of x/(x^2-2) ?

The integral from 4 to 7 of x/(x^2-2) is 1/2 (ln(47)-ln(14))

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integral from 4 to 7 of x/(x^2-2)

\int_{4}^{7} \frac{x}{x ^{2} - 2} d x

\frac{1}{2} (ln (47) - ln (14))

Decimal

0.60554 \dots

Solution steps

\int_{4}^{7} \frac{x}{x ^{2} - 2} d x

Apply u-substitution

: \int_{14}^{47} \frac{1}{2 u} d u

= \int_{14}^{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_{14}^{47} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{14}^{47}

Compute the boundaries:

ln (47) - ln (14)

= \frac{1}{2} (ln (47) - ln (14))

n = 3 \sum \infty \frac{n}{e ^{3 n}}

\frac{d}{d x} (- cos (x^{2}))

x \to 2 lim (6)

x \to 0 lim (\frac{sin ( x ) - x}{x})

n = 1 \sum \infty 4 (\frac{3}{4})^{n}

What is the integral from 4 to 7 of x/(x^2-2) ?
The integral from 4 to 7 of x/(x^2-2) is 1/2 (ln(47)-ln(14))