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

The integral of x/(2x+4) is 1/2 (x+2-2ln(2|x+2|))+C

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integral of x/(2x+4)

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

\frac{1}{2} (x + 2 - 2 ln (2 ∣ x + 2 ∣)) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

\frac{u - 4}{u} : 1 - \frac{4}{u}

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

= \frac{1}{4} (\int 1 d u - \int \frac{4}{u} d u)

\int 1 d u = u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= \frac{1}{4} (u - 4 ln ∣ u ∣)

Substitute back

u = 2 x + 4

= \frac{1}{4} (2 x + 4 - 4 ln ∣ 2 x + 4 ∣)

Simplify

\frac{1}{4} (2 x + 4 - 4 ln ∣ 2 x + 4 ∣) : \frac{1}{2} (x + 2 - 2 ln (2 ∣ x + 2 ∣))

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

Add a constant to the solution

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

x \to 3 lim (\frac{12 - x - 3}{1 + x - x ^{2} + 7})

\frac{d}{d y} (- 2 x y sin (x^{2} y))

\int \frac{1}{sin ^{2} ( θ )} d θ

\int \frac{1}{12 x + 0.02 x ^{2}} d x

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

What is the integral of x/(2x+4) ?
The integral of x/(2x+4) is 1/2 (x+2-2ln(2|x+2|))+C