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

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

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integral of 2/(t-1)

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

2 ln ∣ t - 1 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u} d u

= 2 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 ln ∣ u ∣

Substitute back

u = t - 1

= 2 ln ∣ t - 1 ∣

Add a constant to the solution

= 2 ln ∣ t - 1 ∣ + C

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