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

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

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integral of 2/(1-t^2)

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

ln ∣ t + 1 ∣ - ln ∣ t - 1 ∣ + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \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}

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

Simplify

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

= ln ∣ t + 1 ∣ - ln ∣ t - 1 ∣

Add a constant to the solution

= ln ∣ t + 1 ∣ - ln ∣ t - 1 ∣ + C

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