Hvad er int xln (x) ^ 2?

Hvad er int xln (x) ^ 2?
Anonim

Svar:

Antag at du mener #ln (x) ^ 2 = (LNX) ^ 2 #

Du skal integrere delvist to gange. Svar er:

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

Antag at du mener #ln (x) ^ 2 = ln (x ^ 2) #

Du skal integrere delvist én gang. Svar er:

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

Forklaring:

Antag at du mener #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 ^ annullere (2) / annullere (2) * annullere (2) LNX * 1 / annullere (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 ^ 2/2 (LNX) "dx) = #

# = X ^ 2 / 2ln (x) ^ 2- (x ^ 2 / 2lnx-intx ^ annullere (2) / 2 * 1 / annullere (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 ^ 2/2 (ln (x) ^ 2-LNX + 1/2) + c #

Antag at du mener #ln (x) ^ 2 = ln (x ^ 2) #

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

# 2intxlnxdx = #

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

# = 2 (x ^ 2 / 2lnx-intx ^ 2/2 * (LNX) "dx) = #

# = 2 (x ^ 2 / 2lnx-intx ^ annullere (2) / 2 * 1 / annullere (x) dx) = #

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

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

# = Annullere (2) * x ^ 2 / (annullere (2)) (LNX-1/2) + c = #

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