The integral $$\displaystyle\int_0^1 \cot^{-1}(1 + x + x^2)\,dx$$ is equal to :

$$I = \int_{0}^{1} \cot^{-1}(1 + x + x^2) \, dx = \int_{0}^{1} \tan^{-1}\left(\frac{1}{1 + x + x^2}\right) \, dx$$

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

$$I = \int_{0}^{1} \left[ \tan^{-1}(1 + x) - \tan^{-1}(x) \right] \, dx$$ (Using the identity $$\tan^{-1}\left(\frac{A - B}{1 + AB}\right) = \tan^{-1}(A) - \tan^{-1}(B)$$)

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

For $$\int_{0}^{1} \tan^{-1}(x) \, dx$$:

$$\left[ x \tan^{-1}(x) - \frac{1}{2} \log_e(1 + x^2) \right]_{0}^{1} = \left(1 \cdot \frac{\pi}{4} - \frac{1}{2} \log_e(2)\right) - (0 - 0) = \frac{\pi}{4} - \frac{1}{2}\log_e 2$$

For $$\int_{0}^{1} \tan^{-1}(1 + x) \, dx$$:

Let $$t = 1 + x \implies dt = dx$$. Limits change from $$x=0 \to t=1$$ and $$x=1 \to t=2$$.

$$\int_{1}^{2} \tan^{-1}(t) \, dt = \left[ t \tan^{-1}(t) - \frac{1}{2} \log_e(1 + t^2) \right]_{1}^{2}$$

$$= \left( 2 \tan^{-1}(2) - \frac{1}{2} \log_e(5) \right) - \left( 1 \cdot \frac{\pi}{4} - \frac{1}{2} \log_e(2) \right)$$

$$= 2 \tan^{-1}(2) - \frac{1}{2} \log_e(5) - \frac{\pi}{4} + \frac{1}{2} \log_e(2)$$

$$I = \left( 2 \tan^{-1}(2) - \frac{1}{2} \log_e(5) - \frac{\pi}{4} + \frac{1}{2} \log_e(2) \right) - \left( \frac{\pi}{4} - \frac{1}{2} \log_e(2) \right)$$

$$I = 2 \tan^{-1} 2 - \frac{1}{2} \log_e \frac{5}{4} - \frac{\pi}{2}$$