What is the integral of (4x)/(x+3) ?

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

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integral of (4x)/(x+3)

\int \frac{4 x}{x + 3} d x

4 (x + 3 - 3 ln ∣ x + 3 ∣) + C

Solution steps

\int \frac{4 x}{x + 3} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{x + 3} d x

Apply u-substitution

: \int \frac{u - 3}{u} d u

= 4 \cdot \int \frac{u - 3}{u} d u

Expand

\frac{u - 3}{u} : 1 - \frac{3}{u}

= 4 \cdot \int 1 - \frac{3}{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

= 4 (\int 1 d u - \int \frac{3}{u} d u)

\int 1 d u = u

\int \frac{3}{u} d u = 3 ln ∣ u ∣

= 4 (u - 3 ln ∣ u ∣)

Substitute back

u = x + 3

= 4 (x + 3 - 3 ln ∣ x + 3 ∣)

Add a constant to the solution

= 4 (x + 3 - 3 ln ∣ x + 3 ∣) + C

\int x (x + 1) d x

\int_{1}^{100} \frac{1}{x ^{2}} d x

\frac{d y}{d x}, 2 x = y^{4}

x \to 0 lim (\frac{x}{- 4 x ^{3} - 8 x})

\frac{\partial}{\partial t} (\frac{( s - 3 r )}{t})

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