Key Concepts and Formulas

• Integration by parts: $\int u \, dv = uv - \int v \, du$
• Trigonometric substitution and identities: Using substitutions like $x = t^2$ to simplify integrals.
• $\int \frac{1}{1+x} dx = \tan^{-1}(x) + C$

Step-by-Step Solution

Step 1: Simplify the integrand using trigonometric substitution. We are given the integral $I = \int \sin^{-1}\left(\sqrt{\frac{x}{1+x}}\right) dx$. Let $\theta = \sin^{-1}\left(\sqrt{\frac{x}{1+x}}\right)$. Then, $\sin \theta = \sqrt{\frac{x}{1+x}}$. We want to express $\theta$ in a simpler form, likely involving $\tan^{-1}$.

Step 2: Express $\sin \theta$ in terms of $\tan \theta$. Since $\sin \theta = \sqrt{\frac{x}{1+x}}$, we can construct a right-angled triangle where the opposite side is $\sqrt{x}$ and the hypotenuse is $\sqrt{1+x}$. Then, the adjacent side is $\sqrt{(1+x) - x} = \sqrt{1} = 1$. Thus, $\tan \theta = \frac{\sqrt{x}}{1} = \sqrt{x}$. Therefore, $\theta = \tan^{-1}(\sqrt{x})$.

Step 3: Rewrite the integral. Now, we have $I = \int \tan^{-1}(\sqrt{x}) \, dx$.

Step 4: Apply integration by parts. Let $u = \tan^{-1}(\sqrt{x})$ and $dv = dx$. Then, $du = \frac{1}{1 + (\sqrt{x})^2} \cdot \frac{1}{2\sqrt{x}} \, dx = \frac{1}{2\sqrt{x}(1+x)} \, dx$ and $v = x$. Using integration by parts, $I = x \tan^{-1}(\sqrt{x}) - \int x \cdot \frac{1}{2\sqrt{x}(1+x)} \, dx = x \tan^{-1}(\sqrt{x}) - \frac{1}{2} \int \frac{x}{\sqrt{x}(1+x)} \, dx = x \tan^{-1}(\sqrt{x}) - \frac{1}{2} \int \frac{\sqrt{x}}{1+x} \, dx$

Step 5: Use substitution to solve the remaining integral. Let $t = \sqrt{x}$. Then, $x = t^2$ and $dx = 2t \, dt$. $I = x \tan^{-1}(\sqrt{x}) - \frac{1}{2} \int \frac{t}{1+t^2} (2t \, dt) = x \tan^{-1}(\sqrt{x}) - \int \frac{t^2}{1+t^2} \, dt$

Step 6: Simplify the integral. $I = x \tan^{-1}(\sqrt{x}) - \int \frac{t^2 + 1 - 1}{1+t^2} \, dt = x \tan^{-1}(\sqrt{x}) - \int \left(1 - \frac{1}{1+t^2}\right) \, dt = x \tan^{-1}(\sqrt{x}) - \int 1 \, dt + \int \frac{1}{1+t^2} \, dt$ $I = x \tan^{-1}(\sqrt{x}) - t + \tan^{-1}(t) + C$

Step 7: Substitute back for $t$ and $x$. Substitute $t = \sqrt{x}$ back into the equation: $I = x \tan^{-1}(\sqrt{x}) - \sqrt{x} + \tan^{-1}(\sqrt{x}) + C = (x+1) \tan^{-1}(\sqrt{x}) - \sqrt{x} + C$

Step 8: Identify A(x) and B(x). Comparing this with the given form $A(x) \tan^{-1}(\sqrt{x}) + B(x) + C$, we have $A(x) = x+1$ and $B(x) = -\sqrt{x}$.

Step 9: State the ordered pair (A(x), B(x)). Thus, the ordered pair $(A(x), B(x))$ is $(x+1, -\sqrt{x})$.

Common Mistakes & Tips

• Incorrect Substitution: Choosing the wrong substitution can complicate the integral.
• Forgetting the Constant of Integration: Always remember to add "+ C" to indefinite integrals.
• Algebra Errors: Be careful when simplifying the integral after applying integration by parts.

Summary

We simplified the integral by using trigonometric substitution and integration by parts. We first rewrote the integrand using $\theta = \tan^{-1}(\sqrt{x})$. Then, we applied integration by parts, followed by another substitution to obtain the final result: $(x+1) \tan^{-1}(\sqrt{x}) - \sqrt{x} + C$. Comparing this with the given form, we identified $A(x) = x+1$ and $B(x) = -\sqrt{x}$.

Final Answer The final answer is \boxed{(x + 1, -{\sqrt x })}, which corresponds to option (A).