Ano ang int xln (x) ^ 2?

Sagot:

Kung kaya mo sabihin mo #ln (x) ^ 2 = (lnx) ^ 2 #

Kailangan mong isama sa pamamagitan ng mga bahagi nang dalawang beses. Ang sagot ay:

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

Kung kaya mo sabihin mo #ln (x) ^ 2 = ln (x ^ 2) #

Kailangan mong isama sa pamamagitan ng mga bahagi ng isang beses. Ang sagot ay:

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

Paliwanag:

Kung kaya mo sabihin mo #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 ^ cancel (2) / cancel (2) * cancel (2) lnx * 1 / cancel (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 ^ cancel (2) / 2 * 1 / cancel (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 #

Kung kaya mo sabihin mo #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 ^ cancel (2) / 2 * 1 / cancel (x) dx) = #

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

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

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

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