Question:

<div class='solution'><p>Linear: \(y'+\dfrac{x}{x^2-1}y=\dfrac{x^4+2x}{\sqrt{1-x^2}}\). IF: \(e^{\int x/(x^2-1)\,dx}=e^{\frac{1}{2}\ln|x^2-1|}=\sqrt{|x^2-1|}\). In \((-1,1)\): \(\sqrt{1-x^2}\). So \(d(y\sqrt{1-x^2})/dx=x^4+2x\)... Actually: \(d(y\sqrt{1-x^2})/dx=(x^4+2x)/\sqrt{1-x^2}\cdot\sqrt{1-x^2}=x^4+2x\). \(y\sqrt{1-x^2}=x^5/5+x^2+C\). \(f(0)=0\): \(C=0\). \(f(x)=(x^5/5+x^2)/\sqrt{1-x^2}\). Integral: use odd/even decomposition. \(x^5/5/\sqrt{1-x^2}\) is odd (zero integral on symmetric interval). \(\int_{-a}^{a}x^2/\sqrt{1-x^2}\,dx=2\int_0^a x^2/\sqrt{1-x^2}\,dx\). With \(x=\sin\theta\): \(=2\int_0^{\pi/3}\sin^2\theta\,d\theta=2[\theta/2-\sin2\theta/4]_0^{\pi/3}=\pi/3-\sqrt{3}/4\). <strong>Answer: (1)</strong> \(\pi/3-\sqrt{3}/4\).</p></div>