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 $5(c+d+a b)=$

$$ \int_0^4| | x-2|-x| d x= $$

If $\int_{-a}^a f(x) d x=\int_0^a f(x) d x+\int_0^a g(x) d x$, then $g(x)=$

$f\left(x, y, c_1, c_2\right)=0$ is an equation containing two arbitrary constants $c_1$ and $c_2$. If the differential equation having $f\left(x, y, c_1, c_2\right)=0$ as its general solution is of $k$ th order, then the differential equation corresponding to $x^k+y^k=c^2$ ( $c$ is an arbitrary constant) is