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

The integral from 7/6 to 6 of 1/(1-x^2) is (ln(7)-ln(65))/2

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integral from 7/6 to 6 of 1/(1-x^2)

\int_{\frac{7}{6}}^{6} \frac{1}{1 - x ^{2}} d x

\frac{ln ( 7 ) - ln ( 65 )}{2}

Decimal Notation

- 1.11423 \dots

Solution steps

\int_{\frac{7}{6}}^{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{7}{6}}^{6}

Simplify

[\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{\frac{7}{6}}^{6} : [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{\frac{7}{6}}^{6}

= [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{\frac{7}{6}}^{6}

Compute the boundaries:

\frac{ln ( 7 ) - ln ( 5 ) - ln ( \frac{13}{6} ) - ln ( 6 )}{2}

= \frac{ln ( 7 ) - ln ( 5 ) - ln ( \frac{13}{6} ) - ln ( 6 )}{2}

Simplify

= \frac{ln ( 7 ) - ln ( 65 )}{2}

9 x^{2} cos (x) cot (x)

f (x) = \frac{7}{x}, (4, \frac{7}{2})

\frac{d y}{d x} - 2 y = e^{5 x}

\frac{d}{d x} (- 5 sin (2 x))

\frac{d}{d x} (cos (x + 1))

What is the integral from 7/6 to 6 of 1/(1-x^2) ?
The integral from 7/6 to 6 of 1/(1-x^2) is (ln(7)-ln(65))/2