For `x >1,int sin^(- 1)((2x)/(1+x^2))dx` is equal to

Options

A. `x tan^(-1)x-In|sec(tan^(-1)x)|+c`
B. `x tan^(-1)x+In|sec(tan^(-1)x)|+c`
C. `x tan^(-1)x-In|cos(tan^(-1)x)|+c`
D. none of these

Explanation

To solve the integral \( I = \int \sin^{-1}\left(\frac{2x}{1+x^2}\right) dx \), we can follow these steps:

### Step 1: Substitution
Let \( x = \tan(\theta) \). Then, we have:
\[ dx = \sec^2(\theta) d\theta \]
Also, we can express \( \frac{2x}{1+x^2} \) in terms of \( \theta \):
\[ \frac{2x}{1+x^2} = \frac{2\tan(\theta)}{1+\tan^2(\theta)} = \frac{2\tan(\theta)}{\sec^2(\theta)} = 2\sin(\theta) \]
Thus, we can rewrite the integral as:
\[ I = \int \sin^{-1}(2\sin(\theta)) \sec^2(\theta) d\theta \]

### Step 2: Simplifying the Integral
Using the identity \( \sin^{-1}(2\sin(\theta)) = \theta \) when \( \theta \) is in the appropriate range (which is valid since \( x > 1 \) implies \( \theta \) is in the first quadrant), we have:
\[ I = \int \theta \sec^2(\theta) d\theta \]

### Step 3: Integration by Parts
Let \( u = \theta \) and \( dv = \sec^2(\theta) d\theta \). Then, we have:
\[ du = d\theta \quad \text{and} \quad v = \tan(\theta) \]
Using integration by parts:
\[ I = u v - \int v du = \theta \tan(\theta) - \int \tan(\theta) d\theta \]

### Step 4: Integrating \( \tan(\theta) \)
The integral of \( \tan(\theta) \) is:
\[ \int \tan(\theta) d\theta = -\ln|\cos(\theta)| + C \]
Thus, we can substitute this back into our expression for \( I \):
\[ I = \theta \tan(\theta) + \ln|\cos(\theta)| + C \]

### Step 5: Back Substitution
Recall that \( \theta = \tan^{-1}(x) \) and \( \tan(\theta) = x \). Also, we have:
\[ \cos(\theta) = \frac{1}{\sqrt{1+x^2}} \]
Therefore:
\[ I = \tan^{-1}(x) \cdot x + \ln\left|\frac{1}{\sqrt{1+x^2}}\right| + C \]
This simplifies to:
\[ I = x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2) + C \]

### Final Answer
Thus, the integral \( \int \sin^{-1}\left(\frac{2x}{1+x^2}\right) dx \) is equal to:
\[ I = x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2) + C \]