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)

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

\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

\int \frac{1}{225 + x ^{2}} d x

\int \frac{e ^{x}}{81 - e ^{2 x}} d x

\int x^{2} 16 - x^{2} d x

\int 6 sinh (\frac{x}{2} - ln (3)) d x

y = e^{x}, y = 3, x = 0

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