İnt root3x / (root3x-1) 'in belirsiz integralini nasıl buluyorsunuz?

Cevap:

# (Root3x-1) ^ 3 (+ 9 (root3x-1) ^ 2) / 2 + 9 (root3x-1) + 3LN (abs (root3x-1)) + C #

Açıklama:

Sahibiz #int root3x / (root3x-1) dx #

Vekil # U = (root3x-1) #

# (Du) / (dx) = x ^ (- 2/3) / 3 #

# Dx = 3x ^ (2/3) du #

#int root3x / (root3x-1) (3x ^ (2/3)) du = int (3x) / (root3x-1) du = int (3 (u + 1) ^ 3) / udu = 3int (u ^ 3 + 3u ^ 2 + 3u + 1) / udu = int3u ^ 2 + 9u + 9 + 3 / udu = u ^ 3 (+ 9u ^ 2) / 2 + 9u + 3LN (abs (u)) + C #

Resubstitute # U = root3x-1 #:

# (Root3x-1) ^ 3 (+ 9 (root3x-1) ^ 2) / 2 + 9 (root3x-1) + 3LN (abs (root3x-1)) + C #