Решите определенный интеграл под номером 2 пж

Решите задачу:

$\int_1^2\frac{xdx}{\sqrt{3+2x-x^2}}=\int_1^2\frac{(x-1)dx}{\sqrt{4-(x-1)^2}} + \int_1^2\frac{dx}{\sqrt{4-(x-1)^2}}=\\\\=\int_1^2\frac{d(x-1)^2}{2\sqrt{4-(x-1)^2}}+\int_1^2\frac{d(x-1)}{\sqrt{4-(x-1)^2}}=\\\\=-\frac{1}{2}\int_1^2\frac{d(4-(x-1)^2)}{\sqrt{4-(x-1)^2}}+\int_1^2\frac{d(x-1)}{\sqrt{4-(x-1)^2}}=\\\\=\frac{1}{2}\int_3^4\frac{dt}{\sqrt t}+\int_1^2\frac{d(x-1)}{\sqrt{4-(x-1)^2}}=\\\\=\sqrt{t}|_3^4+\arcsin\frac{x-1}{2}|_1^2=\frac{\pi}{6}+2-\sqrt{3}$