Step 1: Recognise the same pattern but present it as "function times its own derivative":
Since $\tan^{-1}x$ has derivative $\dfrac{1}{1+x^2}$, the integrand is literally $f(x)\cdot f'(x)$ with $f(x)=\tan^{-1}x$. Integrals of this shape always equal $\dfrac{[f(x)]^2}{2}+C$, by reversing the chain rule.
Step 2: Apply this directly as an antiderivative, without a separate substitution step:
$\displaystyle\int \frac{\tan^{-1}x}{1+x^2}\,dx = \frac{(\tan^{-1}x)^2}{2}+C$.
Step 3: Plug in the limits directly:
At $x=1$: $\dfrac{(\pi/4)^2}{2}=\dfrac{\pi^2}{32}$. At $x=0$: $\dfrac{(0)^2}{2}=0$. Subtracting gives $\dfrac{\pi^2}{32}-0$.
Final Answer:
The definite integral equals $\pi^2/32$.
\[ \boxed{\dfrac{\pi^2}{32}} \]