一道积分与极限结合的题求极限 n→∞ lim n ∫上标1+1/n 下标1 ((1+x^n)^0.5)/x dx
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一道积分与极限结合的题求极限 n→∞ lim n ∫上标1+1/n 下标1 ((1+x^n)^0.5)/x dx
一道积分与极限结合的题
求极限 n→∞ lim n ∫上标1+1/n 下标1 ((1+x^n)^0.5)/x dx
一道积分与极限结合的题求极限 n→∞ lim n ∫上标1+1/n 下标1 ((1+x^n)^0.5)/x dx
先算那个积分.
令x^n=tan^2(t),即x=tan^(2/n)(t)
所以∫√(1+x^n)/xdx
=∫(1/cost)/tan^(2/n)(t)*2/n*tan^(2/n-1)(t)*1/cos^2(t)dt
=2/n∫dt/(cos^2(t)sint)
=-2/n∫d(cost)/(cos^2(t)(1-cos^2(t)))
=-2/n∫d(cost)/cos^2(t)-2/n∫d(cost)/(1-cos^2(t))
=2/(ncost)-1/n∫(1/(1+cost)+1/(1-cost))d(cost)
=2/(ncost)-1/nln|(1+cost)/(1-cost)|+C
=2/n*√(1+x^n)-1/nln((√(1+x^n)+1)/(√(1+x^n)-1))+C
所以原式=lim(n→∞)n*[2/n*√(1+x^n)-1/nln((√(1+x^n)+1)/(√(1+x^n)-1))]|(1→1/n)
=lim(n→∞)2(√(1+(1+1/n)^n)-√2)-ln[(√((1+1/n)^n+1)+1)/(√((1+1/n)^n+1)-1)*(√2-1)/(√2+1)]
=2(√(e+1)-√2)-ln[(√(e+1)+1)/(√(e+1)-1)*(√2-1)/(√2+1)]
实在没精力再化简了.