Then $\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{y}{(y+1)(y+4)}$
Write $\frac{y}{(y+1)(y+4)}=\frac{A}{(y+1)}+\frac{B}{(y+4)}$
so that $y=A(y+4)+B(y+1)$
$\therefore \quad A=-\frac{1}{3} $ and $ B=\frac{4}{3}$
$\therefore \mathrm{I}=\int \frac{-\frac{1}{3}}{\left(\mathrm{x}^{2}+1\right)} \mathrm{dx}+\int \frac{4 / 3}{\left(\mathrm{x}^{2}+4\right)} \mathrm{dx}$
$=-\frac{1}{3} \tan ^{-1} \mathrm{x}+\frac{2}{3} \tan ^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{C}$
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$(A)$ $P(E)=\frac{4}{5}, P(F)=\frac{3}{5}$
$(B)$ $P(E)=\frac{1}{5}, P(F)=\frac{2}{5}$
$(C)$ $P(E)=\frac{2}{5}, P(F)=\frac{1}{5}$
$(D)$ $P(E)=\frac{3}{5}, P(F)=\frac{4}{5}$