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

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

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

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

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

Solution steps

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

Complete the square

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

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

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 = v + 1

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

\frac{d}{d x} (- (3 x + 2) cos (2 x))

f (x) = sin (x) + \frac{1}{2} cot (x)

n = 0 \sum \infty \frac{n}{2 n + 4}

\int - \frac{sin ( x ) tan ( x )}{1} d x

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