What is the integral of (4v)/(v^2+4) ?

The integral of (4v)/(v^2+4) is 2ln|v^2+4|+C

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integral of (4v)/(v^2+4)

\int \frac{4 v}{v ^{2} + 4} d v

2 ln v^{2} + 4 + C

Solution steps

\int \frac{4 v}{v ^{2} + 4} d v

Take the constant out:

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

= 4 \cdot \int \frac{v}{v ^{2} + 4} d v

Apply u-substitution

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

= 4 \cdot \int \frac{1}{2 u} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = v^{2} + 4

= 4 \cdot \frac{1}{2} ln v^{2} + 4

Simplify

4 \cdot \frac{1}{2} ln v^{2} + 4 : 2 ln v^{2} + 4

= 2 ln v^{2} + 4

Add a constant to the solution

= 2 ln v^{2} + 4 + C

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What is the integral of (4v)/(v^2+4) ?
The integral of (4v)/(v^2+4) is 2ln|v^2+4|+C