Question

Решите определенный интеграл под номером 2 пж

Answer 1

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

Answer 2

благодарю!