[a,b]int(√(x^2+1) *dx ) =
= b*√(b^2+1)/2 + ln(b+√(b^2+1) )/2 - a*√(a^2+1)/2 - ln(a+√(a^2+1) )/2
Пошаговое объяснение:
[a,b]int(√(x^2+1) *dx )
Замена :
x= ( t- t^-1)/2
Примечание :
(t-t^-1)/2 -монотонно возрастающая функция
dx = dt/2 *( 1 +t^(-2 ) )
x^2+1 = (t^2 -2 +t^-2)/4 +1 = (t^2+2+t^-2)/4 = (t+t^-1)^2 /4
[a,b]int(√(x^2+1) *dx ) = 1/4 * [a,b]int ( dt*(t+t^-1) *(1+t^-2) ) =
= 1/4* [a,b]int ( dt* ( t +2* t^-1 + t^-3 ) = 1/4 [a,b] ( t^2/2 +2*ln(t) - t^-2/2 ) =
= 1/2* [a,b] ( ( (t+t^-1)/2 ) * ( (t-t^-1)/2 ) +ln(t) )
Поскольку :
(t- t^-1)/2 =x
(t+t^-1) /2 =√(x^2+1)
t = x+ √(x^2+1)
[a,b]int(√(x^2+1) *dx ) = [a,b] ( x*√(x^2+1)/2 + ln(x+√(x^2+1) )/2 ) =
[a,b]int(√(x^2+1) *dx ) =
= b*√(b^2+1)/2 + ln(b+√(b^2+1) )/2 - a*√(a^2+1)/2 - ln(a+√(a^2+1) )/2
Пошаговое объяснение:
[a,b]int(√(x^2+1) *dx )
Замена :
x= ( t- t^-1)/2
Примечание :
(t-t^-1)/2 -монотонно возрастающая функция
dx = dt/2 *( 1 +t^(-2 ) )
x^2+1 = (t^2 -2 +t^-2)/4 +1 = (t^2+2+t^-2)/4 = (t+t^-1)^2 /4
[a,b]int(√(x^2+1) *dx ) = 1/4 * [a,b]int ( dt*(t+t^-1) *(1+t^-2) ) =
= 1/4* [a,b]int ( dt* ( t +2* t^-1 + t^-3 ) = 1/4 [a,b] ( t^2/2 +2*ln(t) - t^-2/2 ) =
= 1/2* [a,b] ( ( (t+t^-1)/2 ) * ( (t-t^-1)/2 ) +ln(t) )
Поскольку :
(t- t^-1)/2 =x
(t+t^-1) /2 =√(x^2+1)
t = x+ √(x^2+1)
[a,b]int(√(x^2+1) *dx ) = [a,b] ( x*√(x^2+1)/2 + ln(x+√(x^2+1) )/2 ) =
= b*√(b^2+1)/2 + ln(b+√(b^2+1) )/2 - a*√(a^2+1)/2 - ln(a+√(a^2+1) )/2