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

The integral of 1/(1-x^2) is (ln|x+1|)/2-(ln|x-1|)/2+C

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integral of 1/(1-x^2)

\int \frac{1}{1 - x ^{2}} d x

\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2} + C

Solution steps

\int \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}

Add a constant to the solution

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

x^{2} y^{^{'}} + 8 x y = 17 y^{3}

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

\frac{d}{d x} (1 + ln (x))

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

i = 1 \sum \infty 3 (\frac{1}{4})^{i - 1}

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