Найти интеграл dx / sqrt(8+2х-х^2)

Решите задачу:

$\displaystyle \int \frac{1}{ \sqrt{8+2x-x^2}}\, dx = \int{ \frac{1}{ \sqrt{8+1-1+2x-x^2}}} \, dx=$

$\displaystyle\int { \frac{1}{ \sqrt{9-(x-1)^2}}\,dx = (x-1=a; dx=da)= \int { \frac{1}{ \sqrt{9-a^2}} } \, da=$

$\displaystyle =\int{ \frac{1}{3 \sqrt{1-a^2/9} } \, da= \frac{1}{3} \int{ \frac{1}{ \sqrt{1-a^2/9}}} \, da=(a/3=t;1/3da=dt)=$

$\displaystyle= \frac{1}{3} \int{ \frac{1}{1/3 \sqrt{1-t^2}} } \, dt= \int { \frac{1}{ \sqrt{1-t^2}}} \, dt=arcsin(t)+C=(t=a/3)=$

$\displaystyle=arcsin(a/3)+C=(a=x-1)=arcsin \frac{x-1}{3}+C$