Step 1: Look at the integral we need.
We want $\displaystyle\int_0^1\dfrac{x^2+1}{x^4+1}\,dx$. The given formula for $\int\dfrac{dx}{x^4+1}$ hints at how to handle the denominator.
Step 2: Divide top and bottom by $x^2$.
The integrand becomes $\dfrac{1+1/x^2}{x^2+1/x^2}$. This trick turns the problem into one nice substitution.
Step 3: Choose a smart substitution.
Let $t=x-\dfrac1x$. Then $dt=\left(1+\dfrac{1}{x^2}\right)dx$, which is exactly the numerator. Also $x^2+\dfrac{1}{x^2}=t^2+2$.
Step 4: Rewrite as a standard form.
The integral becomes $\displaystyle\int\dfrac{dt}{t^2+2}=\dfrac{1}{\sqrt2}\tan^{-1}\dfrac{t}{\sqrt2}$. This matches the inverse tangent piece of the given result.
Step 5: Put back $x$ and apply the limits.
So the value is $\dfrac{1}{\sqrt2}\tan^{-1}\dfrac{x-1/x}{\sqrt2}$ from $0$ to $1$, which the question writes as $\dfrac{1}{2\sqrt2}\tan^{-1}\dfrac{\sqrt2 x}{1-x^2}$. As $x\to1$ the inside goes to $+\infty$ so the angle is $\dfrac{\pi}{2}$; at $x=0$ it is $0$.
Step 6: Finish the calculation.
The result is $\dfrac{1}{2\sqrt2}\left(\dfrac{\pi}{2}-0\right)=\dfrac{\pi}{4\sqrt2}$. \[ \boxed{\frac{\pi}{4\sqrt2}} \]