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

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

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

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

2 (x + 1 - ln ∣ x + 1 ∣) + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int \frac{x}{x + 1} d x

Apply u-substitution

: \int \frac{u - 1}{u} d u

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

Expand

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

= 2 \cdot \int 1 - \frac{1}{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

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 2 (u - ln ∣ u ∣)

Substitute back

u = x + 1

= 2 (x + 1 - ln ∣ x + 1 ∣)

Add a constant to the solution

= 2 (x + 1 - ln ∣ x + 1 ∣) + C

x \to 0 lim (\frac{g ( x )}{x})

\frac{d y}{d x} = x e^{x - 2 y}

\int_{1}^{2} (x^{3} + \frac{1}{4 x ^{3}}) d x

\frac{d y}{d x} - \frac{2}{3 x} y + \frac{x ^{2} + 2}{3} y^{4} = 0

\int x \cdot ln (2 x) d x

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