Integral nerijesiv

${\displaystyle \int\frac{x^2\hbox{d}x}{(1+x^2)\sqrt{1-x^2}}=\\ \int\frac{1+x^2-1}{(1+x^2)\sqrt{1-x^2}}\hbox{d}x=\\ \int\frac{\hbox{d}x}{\sqrt{1-x^2}}-\int\frac{\hbox{d}x}{(1+x^2)\sqrt{1-x^2}}}$

Znamo da je $\int\frac{\hbox{d}x}{\sqrt{1-x^2}}=\arcsin x + C$

a $\int\frac{\hbox{d}x}{(1+x^2)\sqrt{1-x^2}}$ rešavamo npr. trećom Ojlerovom smenom:

$\sqrt{1-x^2}=t(x-1)$

Lako nalazimo da je $x=\frac{t^2-1}{t^2+1}$ , odakle sledi:
$1+x^2=\frac{2(t^4+1)}{(t^2+1)^2}$
$\hbox{d}x=\frac{4t}{(t^2+1)^2}\hbox{d}t$
$\sqrt{1-x^2}=\frac{-2t}{t^2+1}$

dakle,

${\displaystyle \int\frac{\hbox{d}x}{(1+x^2)\sqrt{1-x^2}}=\\ \int\frac{\frac{4t}{(t^2+1)^2}}{\frac{2(t^4+1)}{(t^2+1)^2}\cdot \frac{-2t}{t^2+1}}\hbox{d}t=\\ -\int\frac{t^2+1}{t^4+1}\hbox{d}t=\\ -{1\over 2}\int\Bigg (\frac{1}{t^2+1+\sqrt{2}t}+\frac{1}{t^2+1-\sqrt{2}t}\Bigg)\hbox{d}t=\\ -{1\over \sqrt{2}}\Bigg(\newcommand{\arctg}{\mathop{\mathrm{arctg}}}{\arctg(\sqrt{2}t+1)+\arctg(\sqrt{2}t-1)}\Bigg)+D }$

pa ostaje još samo da zamenimo $t=-\sqrt{1+x\over 1-x}$

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