Kaj je int xln (x) ^ 2?

Kaj je int xln (x) ^ 2?
Anonim

Odgovor:

Recimo, da misliš #ln (x) ^ 2 = (lnx) ^ 2 #

Dvakrat se morate vključiti po delih. Odgovor je:

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

Recimo, da misliš #ln (x) ^ 2 = ln (x ^ 2) #

Enkrat se morate vključiti po delih. Odgovor je:

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

Pojasnilo:

Recimo, da misliš #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 ^ preklic (2) / preklic (2) * preklic (2) lnx * 1 / preklic (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 ^ odpoved (2) / 2 * 1 / preklic (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 #

Recimo, da misliš #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 ^ odpoved (2) / 2 * 1 / preklic (x) dx) = #

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

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

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

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