What is the integral from 3 to 5 of x/(x^2-2) ?

The integral from 3 to 5 of x/(x^2-2) is 1/2 (ln(23)-ln(7))

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integral from 3 to 5 of x/(x^2-2)

\int_{3}^{5} \frac{x}{x ^{2} - 2} d x

\frac{1}{2} (ln (23) - ln (7))

Decimal Notation

0.59479 \dots

Solution steps

\int_{3}^{5} \frac{x}{x ^{2} - 2} d x

Apply u-substitution

: \int_{7}^{23} \frac{1}{2 u} d u

= \int_{7}^{23} \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}^{23} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{7}^{23}

Compute the boundaries:

ln (23) - ln (7)

= \frac{1}{2} (ln (23) - ln (7))

\int \frac{2 x + 2}{x ^{2} + 2 x - 1} d x

\int 4 x^{2} sin (π x) d x

\frac{\partial}{\partial x} (ln (2 x^{2} y))

\int \frac{2}{1 + x ^{2}} - \frac{4}{x} d x

y = 25 - x^{2}

What is the integral from 3 to 5 of x/(x^2-2) ?
The integral from 3 to 5 of x/(x^2-2) is 1/2 (ln(23)-ln(7))