What is the integral from 9/8 to 4 of 1/(1-x^2) ?

The integral from 9/8 to 4 of 1/(1-x^2) is (ln(5)-ln(51))/2

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integral from 9/8 to 4 of 1/(1-x^2)

\int_{\frac{9}{8}}^{4} \frac{1}{1 - x ^{2}} d x

\frac{ln ( 5 ) - ln ( 51 )}{2}

Decimal Notation

- 1.16119 \dots

Solution steps

\int_{\frac{9}{8}}^{4} \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}}^{4}

Simplify

[\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{\frac{9}{8}}^{4} : [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{\frac{9}{8}}^{4}

= [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{\frac{9}{8}}^{4}

Compute the boundaries:

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

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

Simplify

= \frac{ln ( 5 ) - ln ( 51 )}{2}

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

\int \frac{1}{3 x 9 x ^{2} - 12} d x

e^{- 1}

\int_{0}^{2} e^{- x} + e d x

\int t e^{- 6 t} d t

What is the integral from 9/8 to 4 of 1/(1-x^2) ?
The integral from 9/8 to 4 of 1/(1-x^2) is (ln(5)-ln(51))/2