Find: ∫ (x^2+1)/((x^2+2)(x^2+4)) d x

$$I=\int \frac{x^{2}+1}{\left(x^{2}+2\right)\left(x^{2}+4\right)} d x $$ Let $\displaystyle x^{2}=y$, then $\displaystyle \frac{x^{2}+1}{\left(x^{2}+2\right)\left(x^{2}+4\right)}=\frac{y+1}{(y+2)(y+4)}$ Let $\displaystyle \frac{y+1}{(y+2)(y+4)}=\frac{A}{y+2}+\frac{B}{y+4}$ this gives $\displaystyle A=-\frac{1}{2}, B=\frac{3}{2}$ $\displaystyle \therefore I=-\frac{1}{2} \int \frac{1}{x^{2}+2} d x+\frac{3}{2} \int \frac{1}{x^{2}+4} d x$ $\displaystyle \Rightarrow I=-\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{3}{4} \tan ^{-1}\left(\frac{x}{2}\right)+c$