The value of $\int_0^1 \tan ^{-1}\left(1-x+x^2\right) \mathrm{d} x$ is

$$ \int_3^5 \frac{\sqrt{x} \mathrm{~d} x}{\sqrt{8-x}+\sqrt{x}}= $$

$$ \int \frac{\mathrm{d} x}{\sqrt{x}+x}= $$

With usual notations, in $\triangle \mathrm{ABC}$, the lengths of two sides are 10 cm and 9 cm respectively. If angles $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are in A.P. then perimeter of $\triangle \mathrm{ABC}$ is