What is the integral from 4/3 to 4 of 1/(1-x^2) ?

The integral from 4/3 to 4 of 1/(1-x^2) is (-2ln(3)+ln(5)-ln(7/3))/2

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integral from 4/3 to 4 of 1/(1-x^2)

\int_{\frac{4}{3}}^{4} \frac{1}{1 - x ^{2}} d x

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

Decimal Notation

- 0.71754 \dots

Solution steps

\int_{\frac{4}{3}}^{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{4}{3}}^{4}

Simplify

[\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{\frac{4}{3}}^{4} : [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{\frac{4}{3}}^{4}

= [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{\frac{4}{3}}^{4}

Compute the boundaries:

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

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

\int (x^{3} + 1)^{2} (3 x) d x

y = \frac{x}{x + 9}, (1, 0.1)

- 2 x^{\frac{2}{3}} - 4, at x = 8

(2 x^{2} - 4)^{3}

x \to 0 - lim (\frac{cot ( π x ) sin ( x )}{2 sec ( x )})

What is the integral from 4/3 to 4 of 1/(1-x^2) ?
The integral from 4/3 to 4 of 1/(1-x^2) is (-2ln(3)+ln(5)-ln(7/3))/2