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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Factor

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

v = u + 1

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

Simplify

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

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

Add a constant to the solution

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

\int t^{2} e^{- 21 t} d t

y^{^{'}} = 3 x y^{2}

\frac{d}{d x} (\frac{x}{3 x ^{2} + 4})

f (x) = - e^{x} x + e^{x}

2 x y^{^{'}} + y = 6 x, y (4) = 22

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