Apakah integral int (1 + e ^ (2x)) ^ (1/2) dx?

Jawapan:

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

Penjelasan:

Pertama kita menggantikan:

# u = e ^ (2x) +1; e ^ (2x) = u-1 #

# (du) / (dx) = 2e ^ (2x); dx = (du) / (2e ^ (2x)) #

du = intsqrt (u) / (2 (u-1)) du = 1 / 2intsqrt (u) / (u-1) du #

Lakukan penggantian kedua:

# v ^ 2 = u; v = sqrt (u) #

# 2v (dv) / (du) = 1; du = 2vdv #

# 1 / 2intv / (v ^ 2-1) 2vdv = intv ^ 2 / (v ^ 2-1) dv = int1 + 1 / (v ^ 2-1) dv #

Split menggunakan pecahan separa:

# 1 / ((v + 1) (v-1)) = A / (v + 1) + B / (v-1) #

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

# v = 1 #:

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

# v = -1 #:

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

Sekarang kita ada:

# -1 / (2 (v + 1)) + 1 / (2 (v-1)) #

dv = int1-1 / (2 (v + 1)) + 1 / (2 (v-1)) dv = 1/2 -ln (abs (v + 1)) + ln (abs (v-1)) + v + C #

Penggantian semula # v = sqrt (u) #:

# 1/2 -ln (abs (sqrt (u) +1)) + ln (abs (sqrt (u) -1)) + sqrt (u)

Penggantian semula # u = 1 + e ^ (2x) #

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