1/(x–1)=t
x–1=1/t
d(x–1)=d(1/t) ⇒ dx=–dt/t2
x–1=1/t ⇒ x=(1/t) +1
1+х–х2=1+(1/t) +1 –((1/t)+1)2 ⇒
1+x–x2=1–(1/t)–(1/t2)=(t2–t–1)/t2
Тогда
∫ dx/(x–1)·√1+x–x2= ∫ (–dt/t2)/ ((1/t)·√t2–t–1/t )=
=– ∫ dt/(t2–t–1)=
выделяем полный квадрат
t2–t+1=(t–(1/2))2–(5/4)
=– ∫ dt/√(t–(1/2))2–(5/4)= – ln|t – (1/2)+√t2–t–1|+С, где
t=1/(x–1)
=–ln|1/(x–1) –(1/2) + √(1/(x–1))2–(1/(x–1))–1|+C=
=–ln|(1/(x–1) – (1/2) + √(1–(x–1)–(x–1)2)/(x–1)|+C=
= –ln|1/(x–1) – (1/2) + √(1+x–x2)/(x–1)|+C