The integral `intxcos^(-1)((1-x^(2))/(1+x^(2)))dx(xgt0)` is equal to

Options

A. `-x+(1+x^(2))tan^(-1)x+C`
B. `x-(1+x^(2))cot^(-1)x+C`
C. `-x+(1+x^(2))cot^(-1)x+C`
D. `x-(1+x^(2))tan^(-1)x+C`

Answer

To solve the integral

\[ I = \int x \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) dx \quad (x > 0) \]

we will use the substitution \( x = \tan \theta \).

### Step 1: Substitution
Let \( x = \tan \theta \). Then, the differential \( dx \) becomes:

\[ dx = \sec^2 \theta \, d\theta \]

### Step 2: Transform the Integral
Now, substitute \( x \) and \( dx \) into the integral:

\[ I = \int \tan \theta \cos^{-1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) \sec^2 \theta \, d\theta \]

Using the identity \( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos 2\theta \), we can rewrite the integral as:

\[ I = \int \tan \theta \cos^{-1} (\cos 2\theta) \sec^2 \theta \, d\theta \]

### Step 3: Simplifying the Integral
Since \( \cos^{-1}(\cos 2\theta) = 2\theta \) (for \( 0 \leq 2\theta \leq \pi \)), we can simplify the integral:

\[ I = \int \tan \theta \cdot 2\theta \cdot \sec^2 \theta \, d\theta \]

### Step 4: Further Simplification
This can be rewritten as:

\[ I = 2 \int \theta \tan \theta \sec^2 \theta \, d\theta \]

### Step 5: Integration by Parts
Let \( u = \theta \) and \( dv = \tan \theta \sec^2 \theta \, d\theta \). Then, we differentiate and integrate:

\[ du = d\theta, \quad v = \int \tan \theta \sec^2 \theta \, d\theta = \frac{1}{2} \tan^2 \theta \]

Using integration by parts:

\[ I = 2 \left( \theta \cdot \frac{1}{2} \tan^2 \theta - \int \frac{1}{2} \tan^2 \theta \, d\theta \right) \]

### Step 6: Integrate \( \tan^2 \theta \)
Recall that \( \tan^2 \theta = \sec^2 \theta - 1 \):

\[ \int \tan^2 \theta \, d\theta = \int (\sec^2 \theta - 1) \, d\theta = \tan \theta - \theta + C \]

### Step 7: Substitute Back
Putting everything together, we have:

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

### Step 8: Final Substitution
Substituting back \( \theta = \tan^{-1}(x) \):

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

Since \( \tan^2(\tan^{-1}(x)) = x^2 \):

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

### Final Result
Thus, the integral evaluates to:

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