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

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

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

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

\frac{- 2 ln ( \frac{3}{4} ) + 2 ln ( \frac{5}{4} )}{2}

Solution steps

\int_{- \frac{1}{4}}^{\frac{1}{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{1}{4}}^{\frac{1}{4}}

Simplify

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

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

Compute the boundaries:

\frac{- 2 ln ( \frac{3}{4} ) + 2 ln ( \frac{5}{4} )}{2}

= \frac{- 2 ln ( \frac{3}{4} ) + 2 ln ( \frac{5}{4} )}{2}

- 9 t^{2}

\int \frac{3 x}{1 + x ^{2}} d x

\int (- 2 x) d x

\frac{\partial}{\partial x} (x + y^{3})

\frac{\partial}{\partial x} (2 y x - 3 y + y^{2})

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