The integral $$\int x \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) dx$$ ($$x > 0$$) is equal to:

We need to evaluate the integral $$\int x \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) dx$$ for $$x > 0$$.

First, recognize that the expression inside the inverse cosine resembles a trigonometric identity. Set $$x = \tan \theta$$, so $$\theta = \tan^{-1} x$$. Since $$x > 0$$, $$\theta$$ lies in $$(0, \pi/2)$$. Then:

$$\frac{1 - x^2}{1 + x^2} = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos 2\theta$$

because $$1 + \tan^2 \theta = \sec^2 \theta$$ and $$\cos 2\theta = \frac{1 - \tan^2 \theta}{\sec^2 \theta} = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$$. Therefore:

$$\cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) = \cos^{-1}(\cos 2\theta)$$

Since $$2\theta \in (0, \pi)$$, and $$\cos^{-1}(\cos \alpha) = \alpha$$ for $$\alpha \in [0, \pi]$$, we have:

$$\cos^{-1}(\cos 2\theta) = 2\theta = 2 \tan^{-1} x$$

Substituting back into the integral:

$$\int x \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) dx = \int x \cdot 2 \tan^{-1} x dx = 2 \int x \tan^{-1} x dx$$

Now, integrate $$\int x \tan^{-1} x dx$$ using integration by parts. Set $$u = \tan^{-1} x$$ and $$dv = x dx$$. Then:

$$du = \frac{1}{1 + x^2} dx, \quad v = \int x dx = \frac{x^2}{2}$$

Integration by parts formula is $$\int u dv = u v - \int v du$$, so:

$$\int x \tan^{-1} x dx = \left(\tan^{-1} x\right) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{1 + x^2} dx = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int \frac{x^2}{1 + x^2} dx$$

Simplify the integrand $$\frac{x^2}{1 + x^2}$$:

$$\frac{x^2}{1 + x^2} = \frac{(1 + x^2) - 1}{1 + x^2} = 1 - \frac{1}{1 + x^2}$$

Thus:

$$\int \frac{x^2}{1 + x^2} dx = \int \left(1 - \frac{1}{1 + x^2}\right) dx = \int 1 dx - \int \frac{1}{1 + x^2} dx = x - \tan^{-1} x$$

Substitute back:

$$\int x \tan^{-1} x dx = \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \left(x - \tan^{-1} x\right) = \frac{x^2}{2} \tan^{-1} x - \frac{x}{2} + \frac{1}{2} \tan^{-1} x$$

Combine the $$\tan^{-1} x$$ terms:

$$= \left(\frac{x^2}{2} + \frac{1}{2}\right) \tan^{-1} x - \frac{x}{2} = \frac{1}{2} (x^2 + 1) \tan^{-1} x - \frac{x}{2}$$

Now, recall the original integral had a factor of 2:

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

Adding the constant of integration:

$$(x^2 + 1) \tan^{-1} x - x + c = -x + (1 + x^2) \tan^{-1} x + c$$

Comparing with the 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. $$x - (1+x^2)\tan^{-1}x + c$$

The result matches option A. Note that $$\cot^{-1} x$$ is not equivalent to $$\tan^{-1} x$$, so options B and C are different.

Hence, the correct answer is Option A.