What is the integral from 3 to 6 of 1/(x^2-1) ?

The integral from 3 to 6 of 1/(x^2-1) is -(ln(7)-ln(10))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 3 to 6 of 1/(x^2-1)

\int_{3}^{6} \frac{1}{x ^{2} - 1} d x

- \frac{ln ( 7 ) - ln ( 10 )}{2}

Decimal Notation

0.17833 \dots

Solution steps

\int_{3}^{6} \frac{1}{x ^{2} - 1} d x

Factor

x^{2} - 1 : - (- x^{2} + 1)

= \int_{3}^{6} \frac{1}{- ( - x ^{2} + 1 )} d x

Take the constant out:

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

= - \int_{3}^{6} \frac{1}{- x ^{2} + 1} d x

Use the common integral:

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

= - [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{3}^{6}

Simplify

- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{3}^{6} : - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{3}^{6}

= - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{3}^{6}

Compute the boundaries:

\frac{ln ( 7 ) - ln ( 5 ) - ln ( 2 )}{2}

= - \frac{ln ( 7 ) - ln ( 5 ) - ln ( 2 )}{2}

Simplify

= - \frac{ln ( 7 ) - ln ( 10 )}{2}

\int x^{\frac{2}{9}} d x

\frac{\partial}{\partial y} (ln (x + 8 y + 5 z))

\frac{d y}{d x} - 2 x y = - 6 x^{2} e^{x^{2}}

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

x \to 0 + lim (x^{x^{3}})

What is the integral from 3 to 6 of 1/(x^2-1) ?
The integral from 3 to 6 of 1/(x^2-1) is -(ln(7)-ln(10))/2