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

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

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integral of-1/(t+1)

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

- ln ∣ t + 1 ∣ + C

Solution steps

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

Apply u-substitution

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

= - \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = t + 1

= - ln ∣ t + 1 ∣

Add a constant to the solution

= - ln ∣ t + 1 ∣ + C

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