What is the integral from 6 to 9 of 1/(x^2-1) ?

The integral from 6 to 9 of 1/(x^2-1) is -(ln(5/4)-ln(7)+ln(5))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 6 to 9 of 1/(x^2-1)

\int_{6}^{9} \frac{1}{x ^{2} - 1} d x

- \frac{ln ( \frac{5}{4} ) - ln ( 7 ) + ln ( 5 )}{2}

Decimal

0.05666 \dots

Solution steps

\int_{6}^{9} \frac{1}{x ^{2} - 1} d x

Factor

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

= \int_{6}^{9} \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_{6}^{9} \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}]_{6}^{9}

Simplify

- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{6}^{9} : - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{6}^{9}

= - [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{6}^{9}

Compute the boundaries:

\frac{ln ( 10 ) - 3 ln ( 2 ) - ln ( 7 ) + ln ( 5 )}{2}

= - \frac{ln ( 10 ) - 3 ln ( 2 ) - ln ( 7 ) + ln ( 5 )}{2}

Simplify

= - \frac{ln ( \frac{5}{4} ) - ln ( 7 ) + ln ( 5 )}{2}

\frac{6}{4 + t}

\int e^{- s x} sin (a x) d x

f (x) = sin (x) ln (6 x)

\int e^{x^{1}} d x

\frac{d}{d x} (\frac{2 x + 5}{3 x - 2})

What is the integral from 6 to 9 of 1/(x^2-1) ?
The integral from 6 to 9 of 1/(x^2-1) is -(ln(5/4)-ln(7)+ln(5))/2