¿Cómo se integra? 1 / (x ^ 2 + 9) ^ (1/2)

# y = int1 / sqrt (x ^ 2 + 9) dx #

poner # x = 3 tant ##rArr t = tan ^ -1 (x / 3) #

Por lo tanto, # 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) / (sect) dt #

# y = int (sect) 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 #

Responder:

Lo sabemos, # int1 / sqrt (X ^ 2 + A ^ 2) dX = ln | X + sqrt (X ^ 2 + A ^ 2) | + c #

Asi que, # I = int1 / (x ^ 2 + 9) ^ (1/2) dx = int1 / sqrt (x ^ 2 + 3 ^ 2) dx #

# => I = ln | x + sqrt (x ^ 2 + 9) | + c #

Explicación:

# II ^ (nd) # Método: Trig. subst.

# I = int1 / (x ^ 2 + 9) ^ (1/2) dx #

Tomar, # x = 3tanu => dx = 3sec ^ 2udu #

# y color (azul) (tanu = x / 3 #

Asi que, # I = int1 / (9tan ^ 2u + 9) ^ (1/2) 3seg ^ 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 #, dónde, #color (azul) (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, donde, C = c-ln3 #