If $$\int \frac{x^3}{\sqrt{1+x^2}} d x=a\left(1+x^2\right)^{\frac{3}{2}}+b \sqrt{1+x^2}+c$$, then $$a+b=$$, (where $$c$$ is constant of integration)

$$\mathrm{A}^{-1}=\frac{-1}{2}\left[\begin{array}{cc}1 & -4 \\ -1 & 2\end{array}\right]$$, then $$2 A+I_2=\quad$$

where $$I_2$$ is a unit matrix of order 2

If $$y=\operatorname{cosec}^{-1}\left[\frac{\sqrt{x}+1}{\sqrt{x}-1}\right]+\cos ^{-1}\left[\frac{\sqrt{x}-1}{\sqrt{x}+1}\right]$$, then $$\frac{d y}{d x}=$$

If $$f(x)=|x-1|+|x-2|+|x-3|, \forall x \in[1,4]$$, then $$\int_\limits1^4 f(x) d x=$$