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

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

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

integral

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

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

Solution steps

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

Factor

t^{2} - 1 : - (- t^{2} + 1)

= \int \frac{1}{- ( - t ^{2} + 1 )} d t

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{- t ^{2} + 1} d t = \frac{ln ∣ t + 1 ∣}{2} - \frac{ln ∣ t - 1 ∣}{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{2}{x ^{2} - 9}

\frac{( x + 3 )}{( x ^{2} + 4 )}

\frac{1}{( 2 x ^{3} )}

sec^{2} (2 x - 1)

y = [3 x + 2] [2 x - 5]

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