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

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

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

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

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

\int_{- \frac{1}{3}}^{\frac{1}{3}} \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}{3}}^{\frac{1}{3}}

Simplify

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

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

Compute the boundaries:

\frac{- 2 ln ( \frac{2}{3} ) + 2 ln ( \frac{4}{3} )}{2}

= \frac{- 2 ln ( \frac{2}{3} ) + 2 ln ( \frac{4}{3} )}{2}

Simplify

= ln (2)

\frac{\partial}{\partial y} (- \frac{y}{x})

\int_{0}^{\frac{π}{3}} cos (\frac{2 x}{3}) d x

x^{2} + 5 x + 1, x^{3} + x^{2} + x + 1

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

\int x^{3} (x^{4} + 16)^{4} d x

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