Cum pot găsi intarctan integral (4x) dx?

Cum pot găsi intarctan integral (4x) dx?
Anonim

Răspuns:

# I = x * tan ^ -1 (4x) -1 / 4log | sqrt (1 + 16x ^ 2) | + C #

# = X * tan ^ -1 (4x) -1 / 8log | (1 + 16x ^ 2) | + C #

Explicaţie:

# (1) I = inttan ^ -1 (4x) dx #

Lăsa, # Tan ^ -1 (4x) = urArr4x = tanurArr4dx = sec ^ 2udu ## RArrdx = 1 / 4sec ^ 2udu #

# I = Intu * 1 / 4sec ^ 2udu = 1 / 4intu * sec ^ 2udu #

Folosind integrarea prin componente, # I = 1/4 u * intsec ^ 2udu-int (d / (du) (u) * intsec ^ 2udu) du = 1/4 u * tanu-int1 * tanudu ## = 1/4 u * tanu-log | Secu | + C ## = 1/4 tan ^ -1 (4x) * (4x) -log | sqrt (1 + tan ^ 2u | + C ## = X * tan ^ -1 (4x) -1 / 4log | sqrt (1 + 16x ^ 2) | + C #

A doua metodă:

# (2) I = int1 * tan ^ -1 (4x) dx ## = Tan ^ -1 (4x) * x-int (1 / (1 + 16x ^ 2) * 4) XDX #

# = X * tan ^ -1 (4x) -1 / 8int (32x) / (1 + 16x ^ 2) dx #

# = X * tan ^ -1 (4x) -1 / 8log | 1 + 16x ^ 2 | + C #