[tex]\displaystyle 1)\,\,\,\int_{0}^1xe^x\, dx = \int_{0}^1x(e^x)'\, dx =xe^x\Big|_{0}^1 - \int_{0}^1(x)'e^x\, dx =\\ \\ = 1\cdot e^1-0\cdot e^0 - \int_{0}^1e^x\, dx =e-(e^x)\big|_{0}^1 = e-e^1+e^0 = \\ \\ = \boxed{1}[/tex]
[tex]\\\displaystyle 2)\,\,\,\int_{0}^{\sqrt{3}}x\,\text{arctg}\,x\, dx = \int_{0}^{\sqrt{3}}\left(\dfrac{x^2}{2}\right)'\,\text{arctg}\,x\, dx=\\ \\ = \dfrac{x^2\,\text{arctg}\,x}{2}\Bigg|_{0}^{\sqrt{3}}-\int_{0}^{\sqrt{3}}\dfrac{x^2}{2}\,(\text{arctg}\,x)'\, dx=\\ \\ =\dfrac{3\,\text{arctg}\,\sqrt{3}}{2}-\dfrac{1}{2}\int_{0}^{\sqrt 3}\dfrac{x^2}{x^2+1}\, dx =\\ \\ =\dfrac{\pi}{2}-\dfrac{1}{2}\int_{0}^{\sqrt{3}}\left(1-\dfrac{1}{x^2+1}\right)\, dx=[/tex]
[tex]\displaystyle =\dfrac{\pi}{2}-\dfrac{x}{2}\Bigg|_{0}^{\sqrt 3}+\dfrac{1}{2}\int_{0}^{\sqrt 3}\dfrac{1}{x^2+1}\,dx =\\=\dfrac{\pi}{2}-\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\text{arctg}\, x\Big|_{0}^{\sqrt{3}}= \dfrac{\pi}{2}-\dfrac{\sqrt{3}}{2}+\dfrac{\pi}{6} =\\ \\ =\boxed{\dfrac{2\pi}{3}-\dfrac{\sqrt{3}}{2}}[/tex]
