1. We need to evaluate $I=\int \sin^{-1}\!\left(\sqrt{\frac{x}{1+x}}\right)dx$ and match it with $I=A(x)\tan^{-1}(\sqrt{x})+B(x)+C.$

2. First simplify the inverse sine expression.

Let $\theta=\tan^{-1}(\sqrt{x}).$ Then $\tan\theta=\sqrt{x}.$ So, $\sin\theta=\frac{\tan\theta}{\sqrt{1+\tan^2\theta}}=\frac{\sqrt{x}}{\sqrt{1+x}}=\sqrt{\frac{x}{1+x}}.$ Hence, $\sin^{-1}\left(\sqrt{\frac{x}{1+x}}\right)=\tan^{-1}(\sqrt{x}).$

Therefore, $I=\int \tan^{-1}(\sqrt{x})\,dx.$

3. Now integrate by parts. Take $u=\tan^{-1}(\sqrt{x}), \qquad dv=dx.$ Then $du=\frac{1}{1+(\sqrt{x})^2}\cdot \frac{1}{2\sqrt{x}}dx=\frac{1}{2\sqrt{x}(1+x)}dx,$ and $v=x.$

So, $I=x\tan^{-1}(\sqrt{x})-\int x\cdot \frac{1}{2\sqrt{x}(1+x)}dx.$ Simplify the integrand: $x\cdot \frac{1}{2\sqrt{x}(1+x)}=\frac{\sqrt{x}}{2(1+x)}.$ Thus, $I=x\tan^{-1}(\sqrt{x})-\frac12\int \frac{\sqrt{x}}{1+x}dx.$

4. Evaluate the remaining integral. Let $t=\sqrt{x}\Rightarrow x=t^2,\quad dx=2t\,dt.$ Then $\int \frac{\sqrt{x}}{1+x}dx=\int \frac{t}{1+t^2}(2t\,dt)=2\int \frac{t^2}{1+t^2}dt.$ Now, $\frac{t^2}{1+t^2}=1-\frac{1}{1+t^2}.$ So, $2\int \frac{t^2}{1+t^2}dt=2\int\left(1-\frac{1}{1+t^2}\right)dt=2(t-\tan^{-1}t).$ Substitute back $t=\sqrt{x}$: $\int \frac{\sqrt{x}}{1+x}dx=2\left(\sqrt{x}-\tan^{-1}(\sqrt{x})\right).$

Therefore, $I=x\tan^{-1}(\sqrt{x})-\frac12\cdot 2\left(\sqrt{x}-\tan^{-1}(\sqrt{x})\right)+C.$ So, $I=x\tan^{-1}(\sqrt{x})-\sqrt{x}+\tan^{-1}(\sqrt{x})+C.$ Hence, $I=(x+1)\tan^{-1}(\sqrt{x})-\sqrt{x}+C.$

5. Compare with $I=A(x)\tan^{-1}(\sqrt{x})+B(x)+C.$ Thus, $A(x)=x+1,\qquad B(x)=-\sqrt{x}.$

6. Match with options:

• A: $(x+1,-\sqrt{x})$ ✅
• B: $(x+1,\sqrt{x})$ ❌
• C: $(x-1,-\sqrt{x})$ ❌
• D: $(x-1,\sqrt{x})$ ❌

So the correct option is A.