Найти длины дуг кривых:
1+(y`)^2=1+(1/x)^2=(x^2+1)/x^2
L= ∫ ^(sqrt(15))_(sqrt(8))sqrt(x^2+1)dx/x=...
Замена переменной:
sqrt(x^2+1)=t
x^2=t^2-1
x=sqrt(t^2-1)
dx=tdt/sqrt(t^2-1)
x=sqrt(15); t=4
x=sqrt(8); t=3
= ∫ ^(4)_(3)t^2dt/(t^2-1)= ∫^(4)_(3)dt+ ∫^(4)_(3)dt/(t^2-1)=
= [b]([/b] t+(1/2)ln|(t-1)/(t+1)|[b])[/b] |^(4)_(3)=(4-3)+(1/2)ln(3/5)-(1/2)ln(2/4)=
=[b]1+(1/2)ln(6/5)[/b]