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

The integral from 0 to x of 1/(1-t^2) is 1/2 (ln|x+1|-ln|x-1|)

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

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

\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)

Solution steps

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

Use the common integral:

\int \frac{1}{1 - t ^{2}} d t = \frac{ln ∣ t + 1 ∣}{2} - \frac{ln ∣ t - 1 ∣}{2}

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

Simplify

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

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

Compute the boundaries:

\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)

= \frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)

\int_{1}^{x} \frac{10}{9 x ^{2}} d x

\int_{x - 3}^{x + 3} y d y

\int_{0}^{3} 53 - (3 x^{2} + 2 x + 20) d x

\int_{1}^{3} \frac{x}{2} d x

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

What is the integral from 0 to x of 1/(1-t^2) ?
The integral from 0 to x of 1/(1-t^2) is 1/2 (ln|x+1|-ln|x-1|)