What is the integral of (8+4x)/((x+2)^2) ?

The integral of (8+4x)/((x+2)^2) is 4ln|x+2|+C

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integral of (8+4x)/((x+2)^2)

\int \frac{8 + 4 x}{( x + 2 ) ^{2}} d x

4 ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{8 + 4 x}{( x + 2 ) ^{2}} d x

Apply u-substitution

: \int \frac{4}{u} d u

= \int \frac{4}{u} d u

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 ln ∣ u ∣

Substitute back

u = x + 2

= 4 ln ∣ x + 2 ∣

Add a constant to the solution

= 4 ln ∣ x + 2 ∣ + C

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What is the integral of (8+4x)/((x+2)^2) ?
The integral of (8+4x)/((x+2)^2) is 4ln|x+2|+C