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Step-by-Step Solution
Step 1: Rewrite the integrand using a trigonometric substitution
We have the integral
$ \displaystyle I = \int \sin^{-1}\!\biggl(\sqrt{\tfrac{x}{1 + x}}\biggr)\,dx $.
Notice that
$ \sin^{-1}\!\biggl(\sqrt{\tfrac{x}{1 + x}}\biggr)
= \sin^{-1}\!\bigl(\tfrac{\sqrt{x}}{\sqrt{1 + x}}\bigr).$
Let
$ \theta = \sin^{-1}\!\bigl(\tfrac{\sqrt{x}}{\sqrt{1 + x}}\bigr). $
Then
$ \sin \theta = \tfrac{\sqrt{x}}{\sqrt{1 + x}}. $
From this, we can find the corresponding tangent:
$ \tan \theta = \tfrac{\sqrt{x}}{1} = \sqrt{x}. $
Hence,
$ \theta = \tan^{-1}(\sqrt{x}). $
Therefore,
$ \displaystyle \sin^{-1}\!\bigl(\sqrt{\tfrac{x}{1 + x}}\bigr)
= \tan^{-1}(\sqrt{x}), $
and the integral simplifies to
$ \displaystyle I = \int \tan^{-1}(\sqrt{x})\,dx. $
Step 2: Apply integration by parts
Let
$ u = \tan^{-1}(\sqrt{x}), \quad dv = dx. $
Then
$ du = \dfrac{1}{1 + (\sqrt{x})^2} \cdot \dfrac{1}{2\sqrt{x}}\,dx
= \dfrac{1}{2\sqrt{x}(1 + x)}\,dx, $
and
$ v = x. $
By the integration by parts formula
$ \int u\,dv = uv - \int v\,du, $
we get
$ \displaystyle I = x \,\tan^{-1}(\sqrt{x})
- \int x \left(\dfrac{1}{2\sqrt{x}(1 + x)}\right)\,dx. $
Step 3: Substitute to simplify the remaining integral
Let
$ t = \sqrt{x} \implies x = t^2 \quad\text{and}\quad dx = 2t\,dt. $
Substituting these into the remaining integral:
$ \begin{aligned}
I &= x\,\tan^{-1}(t)
- \int \dfrac{t^2}{2\,t\,(1 + t^2)} \cdot 2t\,dt \\
&= t^2 \,\tan^{-1}(t)
- \int \dfrac{t^2}{1 + t^2} \,dt.
\end{aligned} $
Step 4: Split and integrate
Observe that
$ \dfrac{t^2}{1 + t^2} = 1 - \dfrac{1}{1 + t^2}, $
because
$ t^2 = (1 + t^2) - 1. $
Therefore,
$ \int \dfrac{t^2}{1 + t^2}\,dt
= \int \Bigl(1 - \dfrac{1}{1 + t^2}\Bigr)\,dt
= \int 1\,dt \;-\; \int \dfrac{1}{1 + t^2}\,dt
= t \;-\; \tan^{-1}(t). $
Thus the integral becomes
$ \displaystyle I = t^2\,\tan^{-1}(t) - \Bigl[\,t - \tan^{-1}(t)\Bigr] + C
= t^2\,\tan^{-1}(t) - t + \tan^{-1}(t) + C. $
Step 5: Substitute back for x
Recall $t = \sqrt{x}$, so
$ t^2 = x $ and $ t = \sqrt{x} $.
Substituting back:
$ \displaystyle I = x \,\tan^{-1}(\sqrt{x})
- \sqrt{x}
+ \tan^{-1}(\sqrt{x})
+ C. $
Combine like terms involving $ \tan^{-1}\!\bigl(\sqrt{x}\bigr) $:
$ \displaystyle I = (x + 1)\,\tan^{-1}(\sqrt{x}) \;-\; \sqrt{x} \;+\; C'. $
Step 6: Identify A(x) and B(x)
Comparing with
$ I = A(x)\,\tan^{-1}(\sqrt{x}) + B(x) + C, $
we see that
$ A(x) = x + 1 \quad\text{and}\quad B(x) = -\sqrt{x}. $
Hence, the ordered pair is
$ \bigl(x + 1,\; -\sqrt{x}\bigr). $
