Найти интеграл dx / sqrt(8+2х-х^2)

[latex]\displaystyle \int \frac{1}{ \sqrt{8+2x-x^2}}\, dx = \int{ \frac{1}{ \sqrt{8+1-1+2x-x^2}}} \, dx= [/latex] [latex]\displaystyle\int { \frac{1}{ \sqrt{9-(x-1)^2}}\,dx = (x-1=a; dx=da)= \int { \frac{1}{ \sqrt{9-a^2}} } \, da= [/latex] [latex]\displaystyle =\int{ \frac{1}{3 \sqrt{1-a^2/9} } \, da= \frac{1}{3} \int{ \frac{1}{ \sqrt{1-a^2/9}}} \, da=(a/3=t;1/3da=dt)= [/latex] [latex]\displaystyle= \frac{1}{3} \int{ \frac{1}{1/3 \sqrt{1-t^2}} } \, dt= \int { \frac{1}{ \sqrt{1-t^2}}} \, dt=arcsin(t)+C=(t=a/3)= [/latex] [latex]\displaystyle=arcsin(a/3)+C=(a=x-1)=arcsin \frac{x-1}{3}+C [/latex]