Given that $\int \frac{1}{x^2+a^2} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c$.

$$ \begin{aligned} & \text { If } \int \frac{1}{x^4+3 x^2+1} d x=a \tan ^{-1}\left(\frac{b\left(x^2-1\right)}{x}\right) \\ & +c \tan ^{-1}\left(\frac{d\left(x^2+1\right)}{x}\right)+k \end{aligned} $$

where $k$ is a constant of integration, then $5(c+d+a b)=$

If $\frac{2 x^2-3 x+5}{(x-7)^3}=\frac{A}{x-7}+\frac{B}{(x-7)^2}+\frac{C}{(x-7)^3}$, then $2 A-3 B+C=$

If $\frac{3 x^2+a x+3}{(2 x+3)\left(x^2+2\right)}=\frac{3}{2 x+3}+\frac{B x+C}{x^2+2}$, then $a(B+C)=$

If $\frac{3 \pi}{4}