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Question Number 173734 by mnjuly1970 last updated on 17/Jul/22

lim_( n→∞) (n +2) ∫_0 ^( 1) x^( n) ln( 1+x )dx=?

limn→∞(n+2)∫01xnln(1+x)dx=?

Answered by mindispower last updated on 17/Jul/22

∫_0 ^1 x^n ln(1+x)dx=I_n =[((x^(n+1) ln(1+x))/(n+1))]_0 ^1 −(1/(n+1))∫_0 ^1 (x^(n+1) /(1+x))dx =((ln(2))/(n+1))−(a_n /(n+1)) (1/2)≤(1/(1+x))≤1,∀x∈[0,1] ⇒(1/(2(n+1)))≤a_n ≤(1/(n+1)) ⇔ −(1/((1+n)^2 ))+((ln(2))/(n+1))≤I_n ≤−(1/(2(1+n)^2 ))+((ln(2))/(n+1)) lim_(n→∞) (n+2)I_n =ln(2)

∫01xnln(1+x)dx=In=[xn+1ln(1+x)n+1]01−1n+1∫01xn+11+xdx=ln(2)n+1−ann+112⩽11+x⩽1,∀x∈[0,1]⇒12(n+1)⩽an⩽1n+1⇔−1(1+n)2+ln(2)n+1⩽In⩽−12(1+n)2+ln(2)n+1limn→∞(n+2)In=ln(2)

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