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

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

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integral of 1/(t^2+t)

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

ln ∣ t ∣ - ln ∣ t + 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{t ^{2} + t} : \frac{1}{t} - \frac{1}{t + 1}

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

\int \frac{1}{t} d t = ln ∣ t ∣

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

= ln ∣ t ∣ - ln ∣ t + 1 ∣

Add a constant to the solution

= ln ∣ t ∣ - ln ∣ t + 1 ∣ + C

\int e^{x} - 1 d x

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