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

The integral of (2t-1)/(t^2+t) is 3ln|t+1|-ln|t|+C

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

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

3 ln ∣ t + 1 ∣ - ln ∣ t ∣ + C

Solution steps

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

Expand

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

\int \frac{2 t}{t ^{2} + t} d t = 2 ln ∣ t + 1 ∣

\int \frac{1}{t ^{2} + t} d t = ln ∣ t ∣ - ln ∣ t + 1 ∣

= 2 ln ∣ t + 1 ∣ - (ln ∣ t ∣ - ln ∣ t + 1 ∣)

Simplify

= 3 ln ∣ t + 1 ∣ - ln ∣ t ∣

Add a constant to the solution

= 3 ln ∣ t + 1 ∣ - ln ∣ t ∣ + C

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What is the integral of (2t-1)/(t^2+t) ?
The integral of (2t-1)/(t^2+t) is 3ln|t+1|-ln|t|+C