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

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

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integral of (t+2)/(t+1)

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

ln ∣ t + 1 ∣ + t + C

Solution steps

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

\frac{t + 2}{t + 1} = \frac{1}{t + 1} + 1

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

Apply the Sum Rule:

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

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

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

\int 1 d t = t

= ln ∣ t + 1 ∣ + t

Add a constant to the solution

= ln ∣ t + 1 ∣ + t + C

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