What is the integral from 6/5 to 4 of 1/(1-x^2) ?

The integral from 6/5 to 4 of 1/(1-x^2) is -(ln(33/5))/2

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integral from 6/5 to 4 of 1/(1-x^2)

\int_{\frac{6}{5}}^{4} \frac{1}{1 - x ^{2}} d x

- \frac{ln ( \frac{33}{5} )}{2}

Decimal Notation

- 0.94353 \dots

Solution steps

\int_{\frac{6}{5}}^{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{6}{5}}^{4}

Simplify

[\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{\frac{6}{5}}^{4} : [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{\frac{6}{5}}^{4}

= [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{\frac{6}{5}}^{4}

Compute the boundaries:

\frac{- ln ( 3 ) - ln ( \frac{11}{5} )}{2}

= \frac{- ln ( 3 ) - ln ( \frac{11}{5} )}{2}

Simplify

= - \frac{ln ( \frac{33}{5} )}{2}

\int x^{\frac{3}{2}} (4 - 4 x) d x

x \to \infty lim (0.72 x)

\frac{d}{d x} (29 - x^{2} - 4 y^{2})

\frac{d}{d x} (4 cos (x) sin (x))

f (x) = \frac{x ^{3} + 9}{x ^{3}}

What is the integral from 6/5 to 4 of 1/(1-x^2) ?
The integral from 6/5 to 4 of 1/(1-x^2) is -(ln(33/5))/2