QPaperGen

We are given:
\[
y = \log (1 + x) - \dfrac{2x}{2 + x}
\]

Differentiate with respect to \(x\):
\[
\dfrac{dy}{dx} = \dfrac{1}{1 + x} - \dfrac{(2 + x)(2) - 2x(1)}{(2 + x)^2}
= \dfrac{1}{1 + x} - \dfrac{4}{(2 + x)^2}
\]

Take LCM and simplify:
\[
\dfrac{dy}{dx} = \dfrac{x^2}{(1 + x)(2 + x)^2}
\]

Now, consider the domain \(x > -1\). Let us examine the sign of the derivative:
\[\] The numerator \(x^2 \geq 0\), and is 0 only when \(x = 0\)
\[\] The denominator \((1 + x)(2 + x)^2 > 0\) for all \(x > -1\), since all terms are positive.

Therefore,
\[
\dfrac{dy}{dx} = \dfrac{x^2}{(1 + x)(2 + x)^2} \geq 0 \text{ for all } x > -1,
\]

\(\textbf{Conclusion:} \)
Since \(\dfrac{dy}{dx} > 0\) for all \(x > -1\), the function is \(\textbf{increasing throughout its domain}\).