What is the integral from 0 to 0.8 of 1/(1-x^2) ?

The integral from 0 to 0.8 of 1/(1-x^2) is 1.09861…

Popular >

integral from 0 to 0.8 of 1/(1-x^2)

\int_{0}^{0.8} \frac{1}{1 - x ^{2}} d x

1.09861 \dots

Solution steps

\int_{0}^{0.8} \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}]_{0}^{0.8}

Simplify

[\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{0}^{0.8} : [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{0}^{0.8}

= [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{0}^{0.8}

Compute the boundaries:

1.09861 \dots

= 1.09861 \dots

\int_{0}^{\frac{π}{2}} 4 sin (x) cos (x) d x

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

\int_{1}^{0} (3 x^{2} + x - 5) d x

\int_{0}^{1} x (3 x^{2} - 4)^{3} d x

\int_{- π}^{π} cos (2 n x) d x

What is the integral from 0 to 0.8 of 1/(1-x^2) ?
The integral from 0 to 0.8 of 1/(1-x^2) is 1.09861…