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

(x + 1)^{2}

a^{12} y^{4} - a^{4} y^{12}

\frac{1}{( x + 4 ) ^{2}}

\frac{36}{( x ^{4} - 4 x + 3 )}

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

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