解ï¼
令y = 1/tãdt = - 1/y² dy
â«(1âx) lnt/(1 + t) dt
= â«(1â1/x) ln(1/y)/(1 + 1/y) * (- 1/y² dy)
= â«(1â1/x) (- lny) * y/(1 + y) * (- 1/y²) dy
= â«(1â1/x) lny/[y(1 + y)] dy
= â«(1â1/x) lnt/[t(1 + t)] dt
f(x) + f(1/x) = â«(1âx) lnt/(1 + t) dt + â«(1â1/x) lnt/(1 + t) dt
= â«(1â1/x) lnt/[t(1 + t)] dt + â«(1â1/x) lnt/(1 + t) dt
= â«(1â1/x) (lnt + t * lnt)/[t(1 + t)] dt
= â«(1â1/x) (1 + t)lnt/[t(1 + t)] dt
= â«(1â1/x) (lnt)/t dt
= â«(1â1/x) lnt dlnt
= (1/2)(lnt)² |[1â1/x]
= (1/2){[ln(1/x)]² - [ln(1)]²}
= (1/2)[- ln(x)]²
= (1/2)(lnx)²
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