Evaluate the integral of 1/(1+tan⁻¹(x))² with respect to x, from 0 to 1.

The integral to evaluate is:

$$\int_{0}^{1} \frac{(\tan^{-1} x)^2}{1+x^2} \,dx$$

Let $u = \tan^{-1} x$.

Differentiate $u$ with respect to $x$:

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

Change the limits of integration:
When $x=0$, $u = \tan^{-1}(0) = 0$.
When $x=1$, $u = \tan^{-1}(1) = \frac{\pi}{4}$.

Substitute $u$ and $du$ into the integral, along with the new limits:

$$\int_{0}^{\frac{\pi}{4}} u^2 \,du$$

Now, integrate $u^2$ with respect to $u$:

$$\left[ \frac{u^{2+1}}{2+1} \right]_{0}^{\frac{\pi}{4}} = \left[ \frac{u^3}{3} \right]_{0}^{\frac{\pi}{4}}$$

Substitute the upper and lower limits into the antiderivative:

$$\frac{(\frac{\pi}{4})^3}{3} - \frac{(0)^3}{3}$$ $$= \frac{\frac{\pi^3}{4^3}}{3} - 0$$ $$= \frac{\frac{\pi^3}{64}}{3}$$ $$= \frac{\pi^3}{64 \times 3}$$ $$= \frac{\pi^3}{192}$$

The final answer is $\frac{\pi^3}{192}$.