= Solution
For $\alpha\ne0$, <integration by substitution> $t=\alpha x$ gives
$$
I(\alpha)=\int_0^1\frac{dx}{1+\alpha^2x^2}=\boxed{\frac{\arctan\alpha}{\alpha}.}
$$
At $\alpha=0$ the integrand is one, so $I(0)=1$, also the continuous <limit> of this expression. For nonzero $\alpha$, <differentiation under the integral sign> is valid on the finite interval and yields
$$
I'(\alpha)=-2\alpha\int_0^1\frac{x^2}{(1+\alpha^2x^2)^2}\,dx
=\frac{\alpha/(1+\alpha^2)-\arctan\alpha}{\alpha^2}.
$$
Set $\alpha=\sqrt3$ and use $\arctan\sqrt3=\pi/3$. The requested value is
$$
\boxed{\int_0^1\frac{x^2}{(1+3x^2)^2}\,dx=-\frac{I'(\sqrt3)}{2\sqrt3}
=\frac{\pi}{18\sqrt3}-\frac1{24}=\frac{\pi\sqrt3}{54}-\frac1{24}.}
$$
Back to article page
