$\int \large\frac{3x^2+5}{(x^2+1)(x^2+4)}$
Put $x^2=-1$ and $x^2=-4$
=> $\int \large\frac{3(-1) +5}{(x^2+1)(-1+4)}$$dx+\int \large\frac{3(-4)+5}{(-4+1)(x^2+4)}$
=> $\large\frac{2}{3} \int \frac{1}{x^2+1}$$dx+\frac{7}{3} \int \large\frac{1}{x^2+4}$$dx$
$\large\frac{2}{3} $$\tan ^{-1}x+\frac{7}{6} \tan ^{-1}(\frac{x}{2})+c$
Hence a is the correct answer.