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

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

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integral of t/(t+1)

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

t + 1 - ln ∣ t + 1 ∣ + C

Solution steps

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

Apply u-substitution

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

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

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= u - ln ∣ u ∣

Substitute back

u = t + 1

= t + 1 - ln ∣ t + 1 ∣

Add a constant to the solution

= t + 1 - ln ∣ t + 1 ∣ + C

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