Hva er int xln (x) ^ 2?

Hva er int xln (x) ^ 2?
Anonim

Svar:

Anta at du mener #ln (x) ^ 2 = (lnx) ^ 2 #

Du må integrere etter deler to ganger. Svaret er:

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

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

Du må integrere etter deler en gang. Svaret er:

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

Forklaring:

Anta 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 ^ avbryt (2) / avbryt (2) * avbryt (2) lnx * 1 / avbryt (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 ^ avbryt (2) / 2 * 1 / avbryt (x) dx) = #

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

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

Anta 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 ^ avbryt (2) / 2 * 1 / avbryt (x) dx) = #

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

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

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

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