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Let $x^2 = y$. Then the integral becomes: $\int \frac{y+1}{(y+2)(2y+1)} dy$ Now, we decompose the fraction into partial fractions: $\frac{y+1}{(y+2)(2y+1)} = \frac{A}{y+2} + \frac{B}{2y+1}$ $y+1 = A(2y+1) + B(y+2)$
To find $A$, let $y = -2$: $-2+1 = A(2(-2)+1) + B(-2+2)$ $-1 = A(-4+1)$ $-1 = -3A$ $A = \frac{1}{3}$ To find $B$, let $y = -\frac{1}{2}$: $-\frac{1}{2} + 1 = A(2(-\frac{1}{2})+1) + B(-\frac{1}{2}+2)$ $\frac{1}{2} = A(0) + B(\frac{3}{2})$ $\frac{1}{2} = \frac{3}{2}B$ $B = \frac{1}{3}$ So, $\frac{y+1}{(y+2)(2y+1)} = \frac{1/3}{y+2} + \frac{1/3}{2y+1}$
Substituting $y = x^2$ back into the expression: $\frac{x^2+1}{(x^2+2)(2x^2+1)} = \frac{1/3}{x^2+2} + \frac{1/3}{2x^2+1}$ Now, we integrate: $\int \frac{x^2+1}{(x^2+2)(2x^2+1)} dx = \frac{1}{3} \int \frac{1}{x^2+2} dx + \frac{1}{3} \int \frac{1}{2x^2+1} dx$ $= \frac{1}{3} \int \frac{1}{x^2+(\sqrt{2})^2} dx + \frac{1}{3} \int \frac{1}{2(x^2+\frac{1}{2})} dx$ $= \frac{1}{3} \int \frac{1}{x^2+(\sqrt{2})^2} dx + \frac{1}{6} \int \frac{1}{x^2+(\frac{1}{\sqrt{2}})^2} dx$ Using the formula $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$: $= \frac{1}{3} \cdot \frac{1}{\sqrt{2}} \tan^{-1}(\frac{x}{\sqrt{2}}) + \frac{1}{6} \cdot \frac{1}{\frac{1}{\sqrt{2}}} \tan^{-1}(\frac{x}{\frac{1}{\sqrt{2}}}) + C$ $= \frac{1}{3\sqrt{2}} \tan^{-1}(\frac{x}{\sqrt{2}}) + \frac{1}{6} \cdot \sqrt{2} \tan^{-1}(\sqrt{2}x) + C$ $= \frac{\sqrt{2}}{6} \tan^{-1}(\frac{x}{\sqrt{2}}) + \frac{\sqrt{2}}{6} \tan^{-1}(\sqrt{2}x) + C$
Final Answer: $\frac{\sqrt{2}}{6} \tan^{-1}(\frac{x}{\sqrt{2}}) + \frac{\sqrt{2}}{6} \tan^{-1}(\sqrt{2}x) + C$
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