What is the integral from 5 to 8 of x/(x^2-2) ?

The integral from 5 to 8 of x/(x^2-2) is 1/2 (ln(62)-ln(23))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 5 to 8 of x/(x^2-2)

\int_{5}^{8} \frac{x}{x ^{2} - 2} d x

\frac{1}{2} (ln (62) - ln (23))

Decimal Notation

0.49582 \dots

Solution steps

\int_{5}^{8} \frac{x}{x ^{2} - 2} d x

Apply u-substitution

: \int_{23}^{62} \frac{1}{2 u} d u

= \int_{23}^{62} \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_{23}^{62} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{23}^{62}

Compute the boundaries:

ln (62) - ln (23)

= \frac{1}{2} (ln (62) - ln (23))

\frac{d}{d x} (\frac{a + x ^{2}}{a - x ^{2}})

3 x ln (1 + x)

\int x^{\frac{1}{3}} + (2 x)^{\frac{1}{3}} d x

y^{^{'}} - 2 y = e^{2 t}

y = 13 - x^{2}, y = x^{2} - 5

What is the integral from 5 to 8 of x/(x^2-2) ?
The integral from 5 to 8 of x/(x^2-2) is 1/2 (ln(62)-ln(23))