If `intsin^(-1)(sqrt(x)/sqrt(x+1))dx=A(x)tan^(-1)sqrt(x)+B(x)+C` then A(x) and B(x) will be

Options

A. `1+x,sqrt(x)`
B. `1-x,-sqrt(x)`
C. `1+x,-sqrt(x)`
D. `1-x,sqrt(x)`

Answer

To solve the problem, we need to find the functions \( A(x) \) and \( B(x) \) in the equation:

\[ \int \sin^{-1}\left(\frac{\sqrt{x}}{\sqrt{x+1}}\right) dx = A(x) \tan^{-1}(\sqrt{x}) + B(x) + C \]

### Step 1: Rewrite the Integral
We start with the integral:

\[ I = \int \sin^{-1}\left(\frac{\sqrt{x}}{\sqrt{x+1}}\right) dx \]

To simplify this, we can use the identity that relates \( \sin^{-1} \) and \( \tan^{-1} \). We can express \( \sin^{-1}(y) \) in terms of \( \tan^{-1} \) using a right triangle.

### Step 2: Construct a Right Triangle
Let \( y = \frac{\sqrt{x}}{\sqrt{x+1}} \). We can construct a right triangle where:
- The opposite side is \( \sqrt{x} \)
- The hypotenuse is \( \sqrt{x+1} \)

Using the Pythagorean theorem, the adjacent side will be:

\[ \sqrt{(\sqrt{x+1})^2 - (\sqrt{x})^2} = \sqrt{(x+1) - x} = \sqrt{1} = 1 \]

Thus, we can write:

\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{x}}{1} = \sqrt{x} \]

This implies:

\[ \sin^{-1}\left(\frac{\sqrt{x}}{\sqrt{x+1}}\right) = \tan^{-1}(\sqrt{x}) \]

### Step 3: Substitute into the Integral
Now, we can rewrite the integral:

\[ I = \int \tan^{-1}(\sqrt{x}) dx \]

### Step 4: Integration by Parts
We will use integration by parts, where we let:

- \( u = \tan^{-1}(\sqrt{x}) \)
- \( dv = dx \)

Then, we differentiate \( u \) and integrate \( dv \):

\[ du = \frac{1}{1+x} \cdot \frac{1}{2\sqrt{x}} dx = \frac{1}{2\sqrt{x}(1+x)} dx \]
\[ v = x \]

Using integration by parts:

\[ I = uv - \int v \, du \]
\[ = x \tan^{-1}(\sqrt{x}) - \int x \cdot \frac{1}{2\sqrt{x}(1+x)} dx \]

### Step 5: Simplify the Remaining Integral
Now we simplify the remaining integral:

\[ \int \frac{x}{2\sqrt{x}(1+x)} dx = \frac{1}{2} \int \frac{\sqrt{x}}{1+x} dx \]

Let \( t = \sqrt{x} \), then \( x = t^2 \) and \( dx = 2t \, dt \):

\[ = \frac{1}{2} \int \frac{t^2}{1+t^2} \cdot 2t \, dt = \int \frac{t^3}{1+t^2} dt \]

### Step 6: Final Integration
Now we can split the integral:

\[ \int \frac{t^3}{1+t^2} dt = \int t dt - \int \frac{t}{1+t^2} dt \]

The first integral is straightforward:

\[ \int t \, dt = \frac{t^2}{2} = \frac{x}{2} \]

The second integral can be solved using substitution:

\[ \int \frac{t}{1+t^2} dt = \frac{1}{2} \ln(1+t^2) \]

### Step 7: Combine Results
Putting everything together, we have:

\[ I = x \tan^{-1}(\sqrt{x}) - \left( \frac{x}{2} - \frac{1}{2} \ln(1+x) \right) + C \]

### Step 8: Identify \( A(x) \) and \( B(x) \)
Now we can express \( I \) in the form:

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

Comparing terms, we find:

\[ A(x) = x \]
\[ B(x) = -\frac{x}{2} + \frac{1}{2} \ln(1+x) \]

### Final Answer
Thus, we conclude:

\[ A(x) = x, \quad B(x) = -\frac{x}{2} + \frac{1}{2} \ln(1+x) \]