`intsin^(-1)((2x)/(1+x^2))dx = intcos^(-1)""(1- x^2)/(1+x^2)dx=inttan^(-1)""(2x)/(1-x^2)dx`

Options

A. `2x tan^(-1) x - log (1+x^2)`
B. `2[x tan^(-1) x - log (1+x^2)]`
C. `2xtan^(-1)x+log(1+x^2)`
D. None of these

Answer

To solve the given problem, we will evaluate the integrals step by step. The question states that:

\[ \int \sin^{-1}\left(\frac{2x}{1+x^2}\right)dx = \int \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx = \int \tan^{-1}\left(\frac{2x}{1-x^2}\right)dx \]

We will show that all three integrals are equal to \(2 \tan^{-1}(x) + C\), where \(C\) is the constant of integration.

### Step 1: Recognize the Integral Forms

The integrals can be recognized as standard forms of inverse trigonometric functions. We will use the fact that:

\[ \int \tan^{-1}(x) \, dx = x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2) + C \]

### Step 2: Integrate \(\tan^{-1}\left(\frac{2x}{1-x^2}\right)\)

Using the substitution \(u = \tan^{-1}(x)\), we can derive the integral for \(\tan^{-1}\left(\frac{2x}{1-x^2}\right)\):

\[ \int \tan^{-1}\left(\frac{2x}{1-x^2}\right)dx \]

Using integration by parts, let \(u = \tan^{-1}(x)\) and \(dv = dx\). Then, \(du = \frac{1}{1+x^2}dx\) and \(v = x\).

Applying integration by parts:

\[ \int u \, dv = uv - \int v \, du \]

This gives us:

\[ x \tan^{-1}(x) - \int x \cdot \frac{1}{1+x^2}dx \]

### Step 3: Solve the Remaining Integral

Now we need to solve the integral:

\[ \int \frac{x}{1+x^2}dx \]

This can be solved using the substitution \(w = 1+x^2\), \(dw = 2x \, dx\), or directly recognizing that:

\[ \int \frac{x}{1+x^2}dx = \frac{1}{2} \ln(1+x^2) + C \]

### Step 4: Combine the Results

Now, substituting back into our integration by parts result:

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

### Step 5: Final Result

Thus, the final result for the integral is:

\[ 2 \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2) + C \]

### Conclusion

The three integrals are equal and can be expressed in terms of \(2 \tan^{-1}(x)\) and the logarithmic term.