Let $I=\int \sin ^2 \sqrt{\frac{x}{a+x}} d x$
Put $x=a \tan ^2 \theta \Rightarrow d x=2 a \tan \theta \sec ^2 \theta d \theta$
$$
\begin{aligned}
& \therefore \quad I=\int \sin \sqrt{\frac{a \tan ^2 \theta}{a+a \tan ^2 \theta}} 2 a \tan \theta \sec ^2 \theta d \theta \\
& =\int \sin ^2 \sqrt{\sin ^2 \theta} \cdot 2 a \tan \theta \sec ^2 \theta d \theta \\
& =2 a \int \theta \tan \theta \sec ^2 \theta d \theta \\
& =2 a\left[\theta \frac{\tan ^2 \theta}{2}-\int \frac{\tan ^2 \theta}{2} \cdot 1 d \theta\right] \\
& =a \theta \tan ^2 \theta-a \int \tan ^2 \theta d \theta \\
& =a \theta \tan ^2 \theta-a(\tan \theta-\theta)+c \\
& =x \tan ^2 \sqrt{\frac{x}{a}}-a \sqrt{\frac{x}{a}}+a \tan ^2 \sqrt{\frac{x}{a}}+c \\
& =(x+a) \tan ^2 \sqrt{\frac{x}{a}}-\sqrt{x a}+\text { constant } \\
& \therefore A(x)=(x+a) \tan ^2 \sqrt{\frac{x}{a}}-\sqrt{x a} \\
&
\end{aligned}
$$