Let I(a, b) = integral from a to b ((1-x²) dx)/((1+x²)

Let $I(a, b) = \int_a^b \frac{(1-x^2) \, dx}{(1+x^2)\sqrt{1+x^4}}$. 1) Show that (i) $I(a, b) = I(-b, -a)$; (ii) $I(a, b) = I(1/a, 1/b)$ ($ab > 0$); (iii) $I(a, 1/a) = 0$. 2) Calculate $I(a, b)$ (first for $a>1,b>1$, using $t=x+1/x$, then use the results from part 1).

Part $1^{\circ}$: Prove the symmetry properties

\subsubsection*{Property (i): $I(a, b) = I(-b, -a)$}

Proof:

Use $u = -x$, so $du = -dx$. When $x=a$: $u=-a$; when $x=b$: $u=-b$.

\[I(a, b) = \int_a^b \frac{(1-x^2) \, dx}{(1+x^2)\sqrt{1+x^4}} = -\int_{-a}^{-b} \frac{(1-u^2) \, du}{(1+u^2)\sqrt{1+u^4}} = \int_{-b}^{-a} \frac{(1-u^2) \, du}{(1+u^2)\sqrt{1+u^4}}\]

Since $u$ is a dummy variable: $\boxed{I(a, b) = I(-b, -a)}$

\subsubsection*{Property (ii): $I(a, b) = I(1/a, 1/b)$ for $ab > 0$}

Proof:

Use $t = \frac{1}{x}$, so $dx = -\frac{1}{t^2} \, dt$. When $x=a$: $t=\frac1a$; when $x=b$: $t=\frac1b$. Express everything in terms of $t$: $1-x^2=\frac{t^2-1}{t^2}$, $1+x^2=\frac{t^2+1}{t^2}$, $1+x^4=\frac{t^4+1}{t^4}$.

\[I(a, b) = \int_{1/a}^{1/b} \frac{\frac{t^2-1}{t^2}}{\frac{t^2+1}{t^2}\sqrt{\frac{t^4+1}{t^4}}} \cdot \left(-\frac{1}{t^2}\right) dt = -\int_{1/a}^{1/b} \frac{t^2-1}{(t^2+1)\sqrt{t^4+1}} \, dt\]

\[= \int_{1/a}^{1/b} \frac{1-t^2}{(1+t^2)\sqrt{1+t^4}} \, dt = I(1/a, 1/b)\]

Therefore: $\boxed{I(a, b) = I(1/a, 1/b)}$ for $ab > 0$.

\subsubsection*{Property (iii): $I(a, 1/a) = 0$}

Proof:

Using Property (ii) with $b=1/a$: $I(a, 1/a) = I(1/a, a)$. But by reversing limits, $I(a, 1/a) = -I(1/a, a)$. Combining:

\[I(a, 1/a) = -I(a, 1/a) \quad\Rightarrow\quad 2I(a, 1/a) = 0 \quad\Rightarrow\quad I(a, 1/a) = 0\]

Therefore: $\boxed{I(a, 1/a) = 0}$

Part $2^{\circ}$: Calculate $I(a, b)$ for $a > 1, b > 1$

Use the substitution $t = x + \frac{1}{x}$

For $x>0$: $dt = \left(1-\frac{1}{x^2}\right)dx$. Also $t^2 = x^2+2+\frac1{x^2}$, so $x^2+\frac1{x^2}=t^2-2$, and $x^4+1=x^2(t^2-2)$, giving $\sqrt{1+x^4}=x\sqrt{t^2-2}$ for $x>1$.

The integrand simplifies (using $\frac{x}{1+x^2}=\frac1t$) to:

\[\frac{1-x^2}{(1+x^2)\sqrt{1+x^4}} \, dx = \frac{-dt}{t\sqrt{t^2-2}}\]

Change limits: when $x=a$: $t=a+\frac1a$; when $x=b$: $t=b+\frac1b$.

\[I(a, b) = -\int_{a+1/a}^{b+1/b} \frac{dt}{t\sqrt{t^2-2}}\]

This is a standard form:

\[\int \frac{dt}{t\sqrt{t^2-2}} = \frac{1}{\sqrt{2}}\arctan\left(\frac{\sqrt{t^2-2}}{\sqrt{2}}\right) + C\]

Therefore, for $a>1, b>1$:

\[\boxed{I(a, b) = \frac{1}{\sqrt{2}}\left[\arctan\left(\frac{\sqrt{(a+1/a)^2-2}}{\sqrt{2}}\right) - \arctan\left(\frac{\sqrt{(b+1/b)^2-2}}{\sqrt{2}}\right)\right]}\]

For general $a, b$, the properties from Part 1 (reflection $x\to-x$ and inversion $x\to 1/x$) extend this result to any signs and magnitudes of $a$ and $b$ with $ab \neq 0$.