With usual notations in $\triangle \mathrm{ABC}$, if $\angle \mathrm{B}=\frac{\pi}{2}$, and $\tan \frac{\mathrm{A}}{2}, \tan \frac{\mathrm{C}}{2}$ are roots of equation $\mathrm{p} x^2+\mathrm{qx}+\mathrm{r}=0$, $\mathrm{p} \neq 0$, then

$\int \frac{\mathrm{d} x}{x\left(x^3+1\right)}=$

If $\tan ^{-1}(x+1)+\tan ^{-1} x+\tan ^{-1}(x-1)=\tan ^{-1} 3$, then for $x<0$ the value of $500 x^4+270 x^2+997=$

Let $\mathrm{f}: \mathbb{R}-\{2\} \rightarrow \mathbb{R}-\{1\}$ defined by $\mathrm{f}(x)=\frac{x-3}{x-2}$ and $\mathrm{g}: \mathbb{R} \rightarrow \mathbb{R}$ defined by $\mathrm{g}(x)=3 x-2$, then sum of all values of $x$ for which $\mathrm{f}^{-1}(x)+\mathrm{g}^{-1}(x)=\frac{19}{6}$ is