What is the integral of (6x)/(x^2+4x+4) ?

The integral of (6x)/(x^2+4x+4) is 6(ln|x+2|+2/(x+2))+C

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integral of (6x)/(x^2+4x+4)

\int \frac{6 x}{x ^{2} + 4 x + 4} d x

6 (ln ∣ x + 2 ∣ + \frac{2}{x + 2}) + C

Solution steps

\int \frac{6 x}{x ^{2} + 4 x + 4} d x

Take the constant out:

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

= 6 \cdot \int \frac{x}{x ^{2} + 4 x + 4} d x

Complete the square

x^{2} + 4 x + 4 : (x + 2)^{2}

= 6 \cdot \int \frac{x}{( x + 2 ) ^{2}} d x

Apply u-substitution

: \int \frac{u - 2}{u ^{2}} d u

= 6 \cdot \int \frac{u - 2}{u ^{2}} d u

Expand

\frac{u - 2}{u ^{2}} : \frac{1}{u} - \frac{2}{u ^{2}}

= 6 \cdot \int \frac{1}{u} - \frac{2}{u ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 6 (\int \frac{1}{u} d u - \int \frac{2}{u ^{2}} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{2}{u ^{2}} d u = - \frac{2}{u}

= 6 (ln ∣ u ∣ - (- \frac{2}{u}))

Substitute back

u = x + 2

= 6 (ln ∣ x + 2 ∣ - (- \frac{2}{x + 2}))

Simplify

= 6 (ln ∣ x + 2 ∣ + \frac{2}{x + 2})

Add a constant to the solution

= 6 (ln ∣ x + 2 ∣ + \frac{2}{x + 2}) + C

\frac{e ^{- 3 s}}{s ^{3} + 4 s}

\frac{\partial}{\partial y} (ln (2 x^{2} + y))

\int \frac{- 3 x ^{3}}{x ^{2} + 9} d x

\frac{d}{d x} (\frac{x y}{z})

x \to 7 lim (2 cos (x))

What is the integral of (6x)/(x^2+4x+4) ?
The integral of (6x)/(x^2+4x+4) is 6(ln|x+2|+2/(x+2))+C