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

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

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integral of 1/((1-t^2))

integral

\frac{1}{( 1 - t ^{2} )}

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

Solution steps

\int \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}

Add a constant to the solution

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

\frac{x}{( x ^{3} + 1 )}

\frac{1}{( x - x ^{2} )}

3 + 2 x

(\frac{x - 1}{x})^{3}

x^{3} - 4 x^{2} + 5 x - 10

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