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

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

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

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

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

Solution steps

\int_{0}^{x} \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}^{x}

Simplify

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

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

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

\int_{1}^{4} \frac{18 ( ln ( x ) ) ^{5}}{x} d x

\int_{0}^{1} \frac{x}{x + 3} d x

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

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