If $$2\int_0^1 \tan^{-1} x\,dx = \int_0^1 \cot^{-1}(1 - x + x^2)\,dx$$, then $$\int_0^1 \tan^{-1}(1 - x + x^2)\,dx$$ is equal to

We are given the equation:

$$2\int_0^1 \tan^{-1} x\,dx = \int_0^1 \cot^{-1}(1 - x + x^2)\,dx$$

We need to find the value of $$\int_0^1 \tan^{-1}(1 - x + x^2)\,dx$$.

First, note that $$1 - x + x^2 = x^2 - x + 1$$. The quadratic $$x^2 - x + 1$$ has discriminant $$(-1)^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 < 0$$, and since the leading coefficient is positive, $$x^2 - x + 1 > 0$$ for all real $$x$$, including in the interval $$[0,1]$$.

Since $$x^2 - x + 1 > 0$$, we can use the identity $$\cot^{-1}(a) = \tan^{-1}\left(\frac{1}{a}\right)$$ for $$a > 0$$. Therefore, the given equation becomes:

$$2\int_0^1 \tan^{-1} x\,dx = \int_0^1 \tan^{-1}\left(\frac{1}{x^2 - x + 1}\right)\,dx$$

Let $$J = \int_0^1 \tan^{-1} x\,dx$$. Then the equation is:

$$2J = \int_0^1 \tan^{-1}\left(\frac{1}{x^2 - x + 1}\right)\,dx$$

Now, let $$I = \int_0^1 \tan^{-1}(x^2 - x + 1)\,dx$$, which is the integral we need to find.

Recall that for any $$a > 0$$, $$\tan^{-1}(a) + \tan^{-1}\left(\frac{1}{a}\right) = \frac{\pi}{2}$$. Applying this with $$a = x^2 - x + 1$$ (which is positive as established), we have:

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

Solving for $$\tan^{-1}(x^2 - x + 1)$$:

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

Therefore, the integral $$I$$ is:

$$I = \int_0^1 \left( \frac{\pi}{2} - \tan^{-1}\left(\frac{1}{x^2 - x + 1}\right) \right) dx = \int_0^1 \frac{\pi}{2} \,dx - \int_0^1 \tan^{-1}\left(\frac{1}{x^2 - x + 1}\right) dx$$

The first integral is:

$$\int_0^1 \frac{\pi}{2} \,dx = \frac{\pi}{2} \times (1 - 0) = \frac{\pi}{2}$$

The second integral is exactly the right-hand side of the given equation, which is $$2J$$. So:

$$I = \frac{\pi}{2} - 2J$$

Now we need to compute $$J = \int_0^1 \tan^{-1} x\,dx$$. We use 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:

$$J = \left[ x \tan^{-1} x \right]_0^1 - \int_0^1 x \cdot \frac{1}{1 + x^2} dx$$

Evaluate the boundary term:

At $$x = 1$$, $$\tan^{-1}(1) = \frac{\pi}{4}$$, so $$1 \cdot \frac{\pi}{4} = \frac{\pi}{4}$$

At $$x = 0$$, $$\tan^{-1}(0) = 0$$, so $$0 \cdot 0 = 0$$

Thus, $$\left[ x \tan^{-1} x \right]_0^1 = \frac{\pi}{4} - 0 = \frac{\pi}{4}$$

Now compute the integral:

$$\int_0^1 \frac{x}{1 + x^2} dx$$

Use substitution. Let $$t = 1 + x^2$$, then $$dt = 2x \, dx$$, so $$x \, dx = \frac{dt}{2}$$.

When $$x = 0$$, $$t = 1$$; when $$x = 1$$, $$t = 2$$.

Thus,

$$\int_0^1 \frac{x}{1 + x^2} dx = \int_1^2 \frac{1}{t} \cdot \frac{dt}{2} = \frac{1}{2} \int_1^2 \frac{dt}{t} = \frac{1}{2} \left[ \ln |t| \right]_1^2 = \frac{1}{2} (\ln 2 - \ln 1) = \frac{1}{2} \ln 2$$

since $$\ln 1 = 0$$.

Therefore,

$$J = \frac{\pi}{4} - \frac{1}{2} \ln 2$$

Now substitute back into the expression for $$I$$:

$$I = \frac{\pi}{2} - 2J = \frac{\pi}{2} - 2 \left( \frac{\pi}{4} - \frac{1}{2} \ln 2 \right) = \frac{\pi}{2} - 2 \cdot \frac{\pi}{4} + 2 \cdot \frac{1}{2} \ln 2 = \frac{\pi}{2} - \frac{\pi}{2} + \ln 2 = \ln 2$$

So, $$\int_0^1 \tan^{-1}(1 - x + x^2)\,dx = \ln 2$$.

Comparing with the options:

A. $$\frac{\pi}{2} + \ln 2$$

B. $$\ln 2$$

C. $$\frac{\pi}{2} - \ln 4$$

D. $$\ln 4$$

Hence, the correct answer is Option B.