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

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

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

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

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

Solution steps

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

Complete the square

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

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

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 = p + 1

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

Simplify

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

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

Add a constant to the solution

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

\int_{0}^{1} (x + 4) (x - 5) d x

\frac{1.6}{3.2 s + 1}

\int \frac{4 x}{1 + 2 x} d x

x = y^{2}, x = y + 6

\frac{\partial}{\partial s} (s cos (t - \frac{π}{2}))

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