Помогите по "математическому анализу".
Вычислим первообразную По формуле Ньютона-Лейбница Так что это ответ под номером 4)
![\int\frac{dx}{\sqrt{3+2x-x^2}} = \int\frac{dx}{\sqrt{4-(x-1)^2}} = \\
=\frac{1}{2}\int\frac{2d[(x-1)/2]}{\sqrt{1-[(x-1)/2]^2}} = \int\frac{dt}{\sqrt{1-t^2}} = \arcsin(t)+C = \arcsin((x-1)/2)+C](https://tex.z-dn.net/?f=%5Cint%5Cfrac%7Bdx%7D%7B%5Csqrt%7B3%2B2x-x%5E2%7D%7D+%3D+%5Cint%5Cfrac%7Bdx%7D%7B%5Csqrt%7B4-%28x-1%29%5E2%7D%7D+%3D+%5C%5C%0A%3D%5Cfrac%7B1%7D%7B2%7D%5Cint%5Cfrac%7B2d%5B%28x-1%29%2F2%5D%7D%7B%5Csqrt%7B1-%5B%28x-1%29%2F2%5D%5E2%7D%7D+%3D+%5Cint%5Cfrac%7Bdt%7D%7B%5Csqrt%7B1-t%5E2%7D%7D+%3D+%5Carcsin%28t%29%2BC+%3D+%5Carcsin%28%28x-1%29%2F2%29%2BC "\int\frac{dx}{\sqrt{3+2x-x^2}} = \int\frac{dx}{\sqrt{4-(x-1)^2}} = \\
=\frac{1}{2}\int\frac{2d[(x-1)/2]}{\sqrt{1-[(x-1)/2]^2}} = \int\frac{dt}{\sqrt{1-t^2}} = \arcsin(t)+C = \arcsin((x-1)/2)+C")
