Step 1: Pick a smarter antiderivative of x:
We again use parts with $u=\tan^{-1}x$ and $dv=x\,dx$. This time we take $v=\frac{x^2+1}{2}$.
This is allowed, because $v$ differs from $\frac{x^2}{2}$ only by a constant. The constant cancels in a definite integral.
Step 2: Write the parts formula:
\[ \int_0^1 x\tan^{-1}x\,dx=\left[\frac{x^2+1}{2}\tan^{-1}x\right]_0^1-\int_0^1\frac{x^2+1}{2}\cdot\frac{1}{1+x^2}\,dx \]
Step 3: Cancel inside the second integral:
The factor $x^2+1$ cancels with $1+x^2$. What is left is $\frac12$.
\[ \int_0^1\frac12\,dx=\frac12 \]
Step 4: Evaluate the boundary term:
At $x=1$ we get $\frac{2}{2}\cdot\frac{\pi}{4}=\frac{\pi}{4}$. At $x=0$ we get $\frac12\cdot 0=0$.
So the boundary term is $\frac{\pi}{4}$.
Step 5: Combine:
\[ \frac{\pi}{4}-\frac12 \]
This matches the value $\frac{\pi}{4}-\frac12\approx 0.285$, which is positive as it must be, since $x\tan^{-1}x\ge 0$ on $[0,1]$.
Final Answer:
The definite integral equals the value below.
\[ \boxed{\dfrac{\pi}{4}-\dfrac{1}{2}} \]