大学极限问题limu^v=lim(1+(u-1))^[1/(u-1)][(u-1)v]=e^lim(u-1)v
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解决时间 2021-12-02 03:07
- 提问者网友：寂寞撕碎了回忆
- 2021-12-01 05:00
大学极限问题limu^v=lim(1+(u-1))^[1/(u-1)][(u-1)v]=e^lim(u-1)v
最佳答案
- 五星知识达人网友：野慌
- 2021-12-01 06:06
u=tan(x/2),x=2arctanu,dx=2/(1+u2)du sinx=2u/(1+u2),cosx=(1-u2)/(1+u2) tanx=2u/(1-u2) ∫1/(1+tanx)dx =∫1/(1+2u/(1-u2))*2/(1+u2)du =2∫(1-u2)/[(1+u2)(1-u2+2u)]du = (1-u2)/[(1+u2)(1-u2+2u)] =(Au+B)/(1+u2)+(Cu+D)/(1-u2+2u) (Au+B)(1-u2+2u)+(Cu+D)(1+u2)=1-u2 (-A+C)u^3+(2A-B+D)u2+(A+2B+C)u+B+D=1-u2 -A+C=0 2A-B+D=-1 A+2B+C=0 B+D=1 A=-1/2,B=1/2,C=-1/2,D=1/2 原式=2∫(-1/2*u+1/2)/(1+u2)du+2*∫(-1/2*u+1/2)/(1-u2+2u)du =∫(1-u)/(1+u2)du+∫(1-u)/(1-u2+2u)du =∫1/(1+u2)du-∫u/(1+u2)du+1/2*∫(-2u+2)/(1-u2+2u)du =arctanu-1/2*ln(1+u2)+1/2*ln(1-u2+2u)+C =arctanu+1/2*ln[(1-u2+2u)/(1+u2)]+C =arctan(tan(x/2))+1/2*ln[(1-tan2(x/2)+2tan(x/2))/(1+tan2(x/2))]+C
全部回答
- 1楼网友：一秋
- 2021-12-01 07:10
u=tan(x/2),x=2arctanu,dx=2/(1+u2)du sinx=2u/(1+u2),cosx=(1-u2)/(1+u2) tanx=2u/(1-u2) ∫1/(1+tanx)dx =∫1/(1+2u/(1-u2))*2/(1+u2)du =2∫(1-u2)/[(1+u2)(1-u2+2u)]du = (1-u2)/[(1+u2)(1-u2+2u)] =(Au+B)/(1+u2)+(Cu+D)/(1-u2+2u) (Au+B)(1-u2+2u)+(Cu+D)(1+u2)=1-u2 (-A+C)u^3+(2A-B+D)u2+(A+2B+C)u+B+D=1-u2 -A+C=0 2A-B+D=-1 A+2B+C=0 B+D=1 A=-1/2,B=1/2,C=-1/2,D=1/2 原式=2∫(-1/2*u+1/2)/(1+u2)du+2*∫(-1/2*u+1/2)/(1-u2+2u)du =∫(1-u)/(1+u2)du+∫(1-u)/(1-u2+2u)du =∫1/(1+u2)du-∫u/(1+u2)du+1/2*∫(-2u+2)/(1-u2+2u)du =arctanu-1/2*ln(1+u2)+1/2*ln(1-u2+2u)+C =arctanu+1/2*ln[(1-u2+2u)/(1+u2)]+C =arctan(tan(x/2))+1/2*ln[(1-tan2(x/2)+2tan(x/2))/(1+tan2(x/2))]+C
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