If $$\int \sin^{-1}\left(\frac{\sqrt{x}}{1+x}\right)dx = A(x)\tan^{-1}(\sqrt{x}) + B(x) + C$$, where C is a constant of integration, then the ordered pair $$(A(x), B(x))$$ can be:

We have to evaluate the integral

$$I=\int \sin^{-1}\!\left(\dfrac{\sqrt{x}}{1+x}\right)\,dx$$

and then express the antiderivative in the form

$$I=A(x)\tan^{-1}(\sqrt{x})+B(x)+C,$$

where $$C$$ is the constant of integration. The task is to identify the ordered pair $$\bigl(A(x),B(x)\bigr)$$ from the four alternatives given.

To handle the inverse-sine function, we introduce the trigonometric substitution

$$x=\tan^{2}\theta\qquad(\theta\ge 0).$$

With this choice

$$\sqrt{x}=\tan\theta ,\qquad 1+x=1+\tan^{2}\theta=\sec^{2}\theta,$$

so that

$$\frac{\sqrt{x}}{1+x}=\frac{\tan\theta}{\sec^{2}\theta}=\tan\theta\cos^{2}\theta =\sin\theta\cos\theta =\frac{\sin 2\theta}{2}.$$

Hence

$$\sin^{-1}\!\left(\frac{\sqrt{x}}{1+x}\right) =\sin^{-1}\!\left(\frac{\sin 2\theta}{2}\right).$$

On the interval $$0\le\theta\le\dfrac{\pi}{2}$$ (which corresponds to $$x\ge 0$$) the principal value of the inverse-sine satisfies

$$\sin^{-1}\!\left(\frac{\sin 2\theta}{2}\right)=\theta.$$

Therefore, after the substitution, the integrand simplifies remarkably:

$$\sin^{-1}\!\left(\frac{\sqrt{x}}{1+x}\right)=\theta.$$

Next we need the differential.

Because $$x=\tan^{2}\theta$$, differentiating both sides gives

$$dx=2\tan\theta\sec^{2}\theta\,d\theta.$$

Putting everything into the integral we get

$$I=\int\theta\,(2\tan\theta\sec^{2}\theta)\,d\theta =2\int\theta\tan\theta\sec^{2}\theta\,d\theta.$$

We integrate this by parts. First, we recognise the standard derivative

$$\frac{d}{d\theta}\!\bigl(\tan\theta\bigr)=\sec^{2}\theta.$$

So we write

$$u=\theta,\quad dv=2\tan\theta\sec^{2}\theta\,d\theta \;\Longrightarrow\; du=d\theta,\quad v=\tan^{2}\theta.$$

Using the integration-by-parts formula

$$\int u\,dv=uv-\int v\,du,$$

we have

$$I=\theta\tan^{2}\theta-\int\tan^{2}\theta\,d\theta.$$

To integrate $$\tan^{2}\theta$$ we employ the identity

$$\tan^{2}\theta=\sec^{2}\theta-1,$$

giving

$$\int\tan^{2}\theta\,d\theta=\int(\sec^{2}\theta-1)\,d\theta =\tan\theta-\theta.$$

Substituting this back,

$$I=\theta\tan^{2}\theta-\bigl(\tan\theta-\theta\bigr) =\theta\bigl(\tan^{2}\theta+1\bigr)-\tan\theta.$$

But $$\tan^{2}\theta+1=\sec^{2}\theta=1+\tan^{2}\theta=1+x,$$ and $$\tan\theta=\sqrt{x}$$. Also, from $$x=\tan^{2}\theta$$ we have $$\theta=\tan^{-1}\!(\sqrt{x})$$. Hence in terms of $$x$$ the antiderivative becomes

$$I=(1+x)\tan^{-1}\!(\sqrt{x})-\sqrt{x}+C.$$

Comparing with the required structure $$I=A(x)\tan^{-1}(\sqrt{x})+B(x)+C,$$ we read off

$$A(x)=x+1,\qquad B(x)=-\sqrt{x}.$$

Among the four alternatives, the ordered pair $$\bigl(A(x),B(x)\bigr)=\bigl(x+1,-\sqrt{x}\bigr)$$ is listed as Option D.

Hence, the correct answer is Option D.