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

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

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integral of x/(2+2x)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u - 2}{4 u} d u

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

Expand

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

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

\int 1 d u = u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

= \frac{1}{4} (u - 2 ln ∣ u ∣)

Substitute back

u = 2 + 2 x

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

Simplify

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

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

Add a constant to the solution

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

x \to 0 lim ((e^{x} - 1)^{x})

\frac{\partial}{\partial x} (x^{- 1} y^{2} + x y^{2} + 3 x y)

\int \frac{x ^{2} - x + 5}{x} d x

n = 1 \sum \infty \frac{n}{n + 5}

\frac{d}{d x} (- \frac{4}{3} sin (x) - \frac{5}{3} cos (x))

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