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

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

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

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

- \frac{1}{2} ln ∣ t + 2 ∣ + \frac{1}{2} ln ∣ t ∣ + C

Solution steps

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

Complete the square

t^{2} + 2 t : (t + 1)^{2} - 1

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

Apply u-substitution

: \int \frac{1}{u ^{2} - 1} d u

= \int \frac{1}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

= \int \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{- u ^{2} + 1} d u

Use the common integral:

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

= - (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = t + 1

= - (\frac{ln ∣ t + 1 + 1 ∣}{2} - \frac{ln ∣ t + 1 - 1 ∣}{2})

Simplify

- (\frac{ln ∣ t + 1 + 1 ∣}{2} - \frac{ln ∣ t + 1 - 1 ∣}{2}) : - \frac{1}{2} ln ∣ t + 2 ∣ + \frac{1}{2} ln ∣ t ∣

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

Add a constant to the solution

= - \frac{1}{2} ln ∣ t + 2 ∣ + \frac{1}{2} ln ∣ t ∣ + C

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