What is the integral of (x-1)/(2x+4) ?

The integral of (x-1)/(2x+4) is 1/2 (x+2-3ln(2|x+2|))+C

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integral of (x-1)/(2x+4)

\int \frac{x - 1}{2 x + 4} d x

\frac{1}{2} (x + 2 - 3 ln (2 ∣ x + 2 ∣)) + C

Solution steps

\int \frac{x - 1}{2 x + 4} d x

Apply u-substitution

: \int \frac{u - 6}{4 u} d u

= \int \frac{u - 6}{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 - 6}{u} d u

Expand

\frac{u - 6}{u} : 1 - \frac{6}{u}

= \frac{1}{4} \cdot \int 1 - \frac{6}{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{6}{u} d u)

\int 1 d u = u

\int \frac{6}{u} d u = 6 ln ∣ u ∣

= \frac{1}{4} (u - 6 ln ∣ u ∣)

Substitute back

u = 2 x + 4

= \frac{1}{4} (2 x + 4 - 6 ln ∣ 2 x + 4 ∣)

Simplify

\frac{1}{4} (2 x + 4 - 6 ln ∣ 2 x + 4 ∣) : \frac{1}{2} (x + 2 - 3 ln (2 ∣ x + 2 ∣))

= \frac{1}{2} (x + 2 - 3 ln (2 ∣ x + 2 ∣))

Add a constant to the solution

= \frac{1}{2} (x + 2 - 3 ln (2 ∣ x + 2 ∣)) + C

\frac{d}{d x} (- \frac{2}{x ^{3}} + \frac{3}{x ^{2}} - 4 x)

\int cos^{4} (4 x) d x

\frac{\partial}{\partial x} (- 2 e^{- 2 x})

\int a^{2} - z^{2} d z

\frac{\partial}{\partial x} (x e^{x} + 2)

What is the integral of (x-1)/(2x+4) ?
The integral of (x-1)/(2x+4) is 1/2 (x+2-3ln(2|x+2|))+C