Step 1: Understanding the Concept:
Integrate using Integration by Parts (ILATE rule): $\int u \cdot v \, dx = u \int v \, dx - \int (u' \int v \, dx) dx$.
Step 2: Formula Application:
Let $u = \tan^{-1} x$ and $v = 1$.
$\int \tan^{-1} x \cdot 1 \, dx = x \tan^{-1} x - \int \frac{x}{1+x^2} \, dx$.
Step 3: Explanation:
The integral $\int \frac{x}{1+x^2} \, dx$ is $\frac{1}{2} \log(1+x^2)$.
Applying limits $[0, 1]$:
$[x \tan^{-1} x - \frac{1}{2} \log(1+x^2)]_0^1$
$= (1 \cdot \frac{\pi}{4} - \frac{1}{2} \log 2) - (0 - 0)$
$= \frac{\pi}{4} - \log(2^{1/2}) = \frac{\pi}{4} - \log \sqrt{2}$.
Step 4: Final Answer:
The integral is $\pi/4 - \log \sqrt{2}$.