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

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

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

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

\frac{ln ( \frac{9}{4} )}{2}

Decimal Notation

0.40546 \dots

Solution steps

\int_{- \frac{1}{5}}^{\frac{1}{5}} \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{1}{5}}^{\frac{1}{5}}

Simplify

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

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

Compute the boundaries:

\frac{- 2 ln ( \frac{4}{5} ) + 2 ln ( \frac{6}{5} )}{2}

= \frac{- 2 ln ( \frac{4}{5} ) + 2 ln ( \frac{6}{5} )}{2}

Simplify

= \frac{ln ( \frac{9}{4} )}{2}

\frac{3 ( y + 1 ) ^{2}}{x ^{4}} d x - \frac{( 2 x ) ( y + 1 )}{x ^{4}} d y = 0

10 \frac{e ^{- 3 s}}{s}

x \to 4 - lim (x - 4)

\int_{0}^{π} (x + sin (x)) d x

a^{5} + cos^{5} (x)

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