[Risolto] Integrale

sqrt (t + 1) = u

t + 1 = u^2

t = u^2 - 1

dt = 2u du

S u * 2u /(u^2 - 1) du = 2 S (u^2 - 1 + 1)/(u^2 - 1) du =

= 2 u + 2 S du/(u^2 - 1)

A/(u - 1) + B/(u + 1)

Au + A + Bu - B = 2

A + B = 0

A - B = 2

A = 1 e B = -1

2u + S ( 1/(u - 1) - 1/(u + 1) ) du =

= 2 u + ln |u - 1| - ln |u + 1| + C =

= 2 sqrt (t + 1) + ln | (sqrt(t + 1) - 1)/(sqrt (t + 1) + 1) | + C =

= 2 sqrt (t + 1) + ln ( sqrt (t + 1) - 1)^2 / t | + C

* f(t) = √(t + 1)/t
* F(t) = ∫ f(t)*dt = ∫ (√(t + 1)/t)*dt
Con
* u = √(t + 1) ≡ t = u^2 - 1
* du = (1/(2*√(t + 1)))*dt ≡ dt = (2*√(t + 1))*du = 2*u*du
si ha
* ∫ (√(t + 1)/t)*dt = ∫ (u/(u^2 - 1))*2*u*du = 2*∫ (u^2)*du/(u^2 - 1)
* u^2/(u^2 - 1) = u^2/(u^2 - 1) = 1/(2*(u - 1)) - 1/(2*(u + 1)) + 1
* ∫ (u^2)*du/(u^2 - 1) = ∫ du/(2*(u - 1)) - ∫ du/(2*(u + 1)) + ∫ du
da qui in poi te la cavi da te, vero?