What is the integral from 9/8 to 6 of 1/(1-x^2) ?

The integral from 9/8 to 6 of 1/(1-x^2) is (ln(7)-ln(85))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 9/8 to 6 of 1/(1-x^2)

\int_{\frac{9}{8}}^{6} \frac{1}{1 - x ^{2}} d x

\frac{ln ( 7 ) - ln ( 85 )}{2}

Decimal

- 1.24837 \dots

Solution steps

\int_{\frac{9}{8}}^{6} \frac{1}{1 - x ^{2}} d x

Use the common integral:

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

= [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{\frac{9}{8}}^{6}

Simplify

[\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{\frac{9}{8}}^{6} : [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{\frac{9}{8}}^{6}

= [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{\frac{9}{8}}^{6}

Compute the boundaries:

\frac{ln ( 7 ) - ln ( 5 ) - ln ( \frac{17}{8} ) - 3 ln ( 2 )}{2}

= \frac{ln ( 7 ) - ln ( 5 ) - ln ( \frac{17}{8} ) - 3 ln ( 2 )}{2}

Simplify

= \frac{ln ( 7 ) - ln ( 85 )}{2}

\frac{\partial}{\partial x} (\frac{( x - y )}{x y + 3})

\int cos^{4} (8 x) d x

x \to 0 lim (\frac{3 x ^{2} + 5 x}{x})

y^{^{''}} - y^{^{'}} - 2 y = e^{2 x}

\int x^{3} x - 2 d x

What is the integral from 9/8 to 6 of 1/(1-x^2) ?
The integral from 9/8 to 6 of 1/(1-x^2) is (ln(7)-ln(85))/2