# y = int1 / sqrt (x ^ 2 + 9) dx #
posar # x = 3 tant ##rArr t = tan ^ -1 (x / 3) #
Per tant, # dx = 3sec ^ 2tdt #
# y = int (3sec ^ 2t) / sqrt (9tan ^ 2t + 9) dt #
# y = int (sec ^ 2t) / sqrt (tan ^ 2t + 1) dt #
# y = int (sec ^ 2t) / sqrt (sec ^ 2t) dt #
# y = int (sec ^ 2t) / (secta) dt #
# y = int (secta) dt #
# y = ln | sec t + tan t | + C #
# y = ln | sec (tan ^ -1 (x / 3)) + tan (tan ^ -1 (x / 3)) | + C #
# y = ln | sec (tan ^ -1 (x / 3)) + x / 3) + C #
# y = ln | sqrt (1 + x ^ 2/9) + x / 3 | + C #
Resposta:
Ho sabem, # int1 / sqrt (X ^ 2 + A ^ 2) dX = ln | X + sqrt (X ^ 2 + A ^ 2) | + c #
Tan, # I = int1 / (x ^ 2 + 9) ^ (1/2) dx = int1 / sqrt (x ^ 2 + 3 ^ 2) dx #
# => I = ln | x + sqrt (x ^ 2 + 9) | + c #
Explicació:
# II ^ (nd) # mètode: Trig. subst.
# I = int1 / (x ^ 2 + 9) ^ (1/2) dx #
Prengui, # x = 3tanu => dx = 3sec ^ 2udu #
# i color (blau) (tanu = x / 3 #
Tan, # I = int1 / (9tan ^ 2u + 9) ^ (1/2) 3sec ^ 2udu #
# = int (3sec ^ 2u) / ((9sec ^ 2u) ^ (1/2)) du #
# = int (3sec ^ 2u) / (3secu) du #
# = intsecudu #
# = ln | secu + tanu | + c #
# = ln | sqrt (tan ^ 2u + 1) + tanu | + c #, on, #color (blau) (tanu = x / 3 #
#:. I = ln | sqrt (x ^ 2/9 + 1) + x / 3 | + c #
# = ln | sqrt (x ^ 2 + 9) / 3 + x / 3 | + c #
# = ln | (sqrt (x ^ 2 + 9) + x) / 3 | + c #
# = ln | sqrt (x ^ 2 + 9) + x | -ln3 + c #
# = ln | x + sqrt (x ^ 2 + 9) | + C, on, C = c-ln3 #
