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

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

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integral of 2/(t^2-1)

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

- ln ∣ t + 1 ∣ + ln ∣ t - 1 ∣ + C

Solution steps

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

Factor

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

= 2 \cdot \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

= 2 (- \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}

= 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

\int \frac{- 2}{x ^{2} - 25} d x

\int sin (x) 2 + 3 cos (x) d x

y = (3 x^{2} + 2) (2 x - 5)

\frac{d}{d y} (\frac{y ^{4}}{8} + \frac{1}{4 y ^{2}})

f (x) = \frac{36}{( 1 + 3 x ) ^{3}}

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