Step 1: Find where the function lives.
$f(x) = \log(1+x) - \dfrac{2x}{2+x}$. The log needs $1 + x > 0$, so $x > -1$. The domain is $(-1, \infty)$, so any interval outside this is out.
Step 2: Recall the increasing test.
A function increases where its derivative is positive, so we compute $f'(x)$ and check its sign.
Step 3: Differentiate.
The log gives $\dfrac{1}{1+x}$. For the fraction use the quotient rule:
\[ \frac{d}{dx}\left(\frac{2x}{2+x}\right) = \frac{2(2+x) - 2x}{(2+x)^2} = \frac{4}{(2+x)^2} \]
So $f'(x) = \dfrac{1}{1+x} - \dfrac{4}{(2+x)^2}$.
Step 4: Set the derivative positive.
\[ \frac{1}{1+x} > \frac{4}{(2+x)^2} \;\Rightarrow\; (2+x)^2 > 4(1+x) \]
(valid since $1+x > 0$ on the domain).
Step 5: Simplify the inequality.
\[ x^2 + 4x + 4 > 4 + 4x \;\Rightarrow\; x^2 > 0 \]
This is true for every $x$ except $x = 0$, where it is zero.
Step 6: Conclude.
On the domain $x > -1$, the derivative is positive everywhere except at the single point $x = 0$ where it is zero. So $f$ is increasing on the whole domain $(-1, \infty)$.
\[ \boxed{(-1, \infty)} \]