Nasıl bütünleşirsiniz? 1 / (x ^ 2 + 9) ^ (1/2)

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

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

Bu nedenle, # dx = 3sn ^ 2dpt #

# y = int (3sn ^ 2t) / sqrt (9tan ^ 2t + 9) dt #

# y = int (sn ^ 2t) / sqrt (tan ^ 2t + 1) dt #

# y = int (sn ^ 2t) / sqrt (sn ^ 2t) dt #

# y = int (sn ^ 2t) / (sect) dt #

# y = int (tarikat) dt #

# y = ln | sn t + tan t | + C #

# y = ln | sn (tan ^ -1 (x / 3)) + tan (tan ^ -1 (x / 3)) | + C #

# y = ln | sn (tan ^ -1 (x / 3)) + x / 3) | + C #

# y = ln | sqrt (1 + x ^ 2/9) + x / 3 | + C #

Cevap:

Biz biliyoruz ki, # int1 / sqrt (X ^ 2 + A ^ 2) dX = ln | X + sqrt (X ^ 2 + A ^ 2) | + c #

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

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

Açıklama:

# II ^ (nd) # yöntem: Trig. subst.

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

Take # X = 3tanu => dx = 3sec ^ 2udu #

#ve renk (mavi) (tanu = x / 3 #

Yani, # I = int1 / (9tan ^ 2u + 9) ^ (1/2) 3 saniye ^ 2udu #

# = İnt (3 sn ^ 2u) / ((9sec ^ 2u) ^ (1/2)) du #

# = İnt (3sec ^ 2u) / (3secu) du #

# = İntsecudu #

# = Ln | secu + tanu | + c #

# = Ln | sqrt (tan ^ 2u + 1) tanu | + c #, nerede, #color (mavi) (tanu = X / 3 #

#:. ı = İn | sqrt (x ^ 2/9 + 1) x / 3 | + c #

# = Ln | sqrt (x ^ 2 + 9) / 3 + x / 3 | + c #

# = İn (| sqrt (x ^ 2 + 9) + X) / 3 | + c #

# = Ln | sqrt (x ^ 2 + 9) + x | -ln3 + c #

# = ln | x + sqrt (x ^ 2 + 9) | + C, burada, C = c-ln3 #