Integration gamit ang substitution intsqrt (1 + x ^ 2) / x dx? Paano ko malutas ang tanong na ito, maaring tulungan ako?

Sagot:

(abs (sqrt (1 + x ^ 2) +1)) + 1 / 2ln (abs (sqrt (1 + x ^ 2) -1)) + C #

Paliwanag:

Gamitin # u ^ 2 = 1 + x ^ 2 #, # x = sqrt (u ^ 2-1) #

# 2u (du) / (dx) = 2x #, # dx = (udu) / x #

#intsqrt (1 + x ^ 2) / xdx = int (usqrt (1 + x ^ 2)) / x ^ 2du #

# intu ^ 2 / (u ^ 2-1) du = int1 + 1 / (u ^ 2-1) du #

# 1 / (u ^ 2-1) = 1 / ((u + 1) (u-1)) = A / (u + 1) + B / (u-1) #

# 1 = A (u-1) + B (u + 1) #

# u = 1 #

# 1 = 2B #, # B = 1/2 #

# u = -1 #

# 1 = -2A #, # A = -1 / 2 #

# int1-1 / (2 (u + 1)) + 1 / (2 (u-1)) du = u-1 / 2ln (abs (u + 1)) + 1 / 2ln (abs (u-1)) + C #

Paglalagay # u = sqrt (1 + x ^ 2) # bumalik sa ay nagbibigay sa:

(abs (sqrt (1 + x ^ 2) +1)) + 1 / 2ln (abs (sqrt (1 + x ^ 2) -1)) + C #