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Let $x^2 = y$. Then the integrand becomes $\frac{y+1}{(y+2)(y+4)}$. We can decompose this into partial fractions:
$$\frac{y+1}{(y+2)(y+4)} = \frac{A}{y+2} + \frac{B}{y+4}$$
Multiplying both sides by $(y+2)(y+4)$, we get: $$y+1 = A(y+4) + B(y+2)$$ To find $A$, let $y = -2$: $$-2+1 = A(-2+4) + B(0) \implies -1 = 2A \implies A = -\frac{1}{2}$$ To find $B$, let $y = -4$: $$-4+1 = A(0) + B(-4+2) \implies -3 = -2B \implies B = \frac{3}{2}$$
So, we have: $$\frac{y+1}{(y+2)(y+4)} = -\frac{1}{2(y+2)} + \frac{3}{2(y+4)}$$ Substituting back $y = x^2$, we get: $$\frac{x^2+1}{(x^2+2)(x^2+4)} = -\frac{1}{2(x^2+2)} + \frac{3}{2(x^2+4)}$$
Now, we integrate: $$\int \frac{x^2+1}{(x^2+2)(x^2+4)} dx = \int \left(-\frac{1}{2(x^2+2)} + \frac{3}{2(x^2+4)}\right) dx$$ $$= -\frac{1}{2} \int \frac{1}{x^2+2} dx + \frac{3}{2} \int \frac{1}{x^2+4} dx$$ We know that $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$. So, $$= -\frac{1}{2} \cdot \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x}{\sqrt{2}}\right) + \frac{3}{2} \cdot \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + C$$ $$= -\frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{x}{\sqrt{2}}\right) + \frac{3}{4} \tan^{-1}\left(\frac{x}{2}\right) + C$$
Final Answer: $-\frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{x}{\sqrt{2}}\right) + \frac{3}{4} \tan^{-1}\left(\frac{x}{2}\right) + C$
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