Wat is int xln (x) ^ 2?

Wat is int xln (x) ^ 2?
Anonim

Antwoord:

Stel dat je het meent #ln (x) ^ 2 = (lnx) ^ 2 #

Je moet twee keer per onderdeel integreren. Antwoord is:

# ^ X 2/2 (ln (x) ^ 2-lnx + 1/2) + c #

Stel dat je het meent #ln (x) ^ 2 = ln (x ^ 2) #

Je moet één keer per onderdeel integreren. Antwoord is:

# X ^ 2 (LNX-1/2) + c #

Uitleg:

Stel dat je het meent #ln (x) ^ 2 = (lnx) ^ 2 #

#intxln (x) = ^ 2dx #

# = Int (x ^ 2/2) 'ln (x) = ^ 2dx #

# = X ^ 2 / 2ln (x) ^ 2-INTX ^ 2/2 (ln (x) ^ 2) dx = #

# = X ^ 2 / 2ln (x) ^ 2-INTX ^ annuleren (2) / uitschakelen (2) * annuleren (2) lnx * 1 / annuleren (x) dx = #

# = X ^ 2 / 2ln (x) ^ 2-intxlnxdx = #

# = X ^ 2 / 2ln (x) ^ 2-int (x ^ 2/2) = lnxdx #

# = X ^ 2 / 2ln (x) ^ 2- (x ^ 2 / 2lnx-INTX ^ 02/02 (lnx) dx) = #

# = X ^ 2 / 2ln (x) ^ 2- (x ^ 2 / 2lnx-INTX ^ annuleren (2) / 2 * 1 / annuleren (x) dx) = #

# = X ^ 2 / 2ln (x) ^ 2- (x ^ 2 / 2lnx-1 / 2intxdx) = #

# = X ^ 2 / 2ln (x) ^ 2- (x ^ 2 / 2lnx-1 / 2x ^ 2/2) + c = #

# = X ^ 2 / 2ln (x) ^ 2- (x ^ 2 / 2lnx-x ^ 2/4) + c = #

# = X ^ 2 / 2ln (x) ^ 2-x ^ 2 / 2lnx + x ^ 2/4 + c = #

# = X ^ 02/02 (ln (x) ^ 2-lnx + 1/2) + c #

Stel dat je het meent #ln (x) ^ 2 = ln (x ^ 2) #

#intxln (x) = ^ 2dx INTX * 2lnxdx #

# 2intxlnxdx = #

# = 2INT (x ^ 02/02) lnxdx = #

# = 2 (x ^ 2 / 2lnx-INTX ^ 2/2 * (lnx) dx) = #

# = 2 (x ^ 2 / 2lnx-INTX ^ annuleren (2) / 2 * 1 / annuleren (x) dx) = #

# = 2 (x ^ 2 / 2lnx-1 / 2intxdx) = #

# = 2 (x ^ 2 / 2lnx-1 / 2x ^ 2/2) + c = #

# = Annuleren (2) * x ^ 2 / (annuleren (2)) (LNX-1/2) + c = #

# = X ^ 2 (LNX-1/2) + c #