求教这几道微积分题目怎么做？

Question

想要详细的解题过程。题目如图。非常感谢

Answer 1

1
∫[e^t+e^(-t)-2]dt=e^x-e^(-x)-2x
lim(x→0) ∫[0,x] [e^t+e^(-t)]dt /(1-cosx)=lim(x→0) [e^x-e^(-x)-2x]/(1-cosx)
=lim(x→0) [e^x+e^(-x)-2]/sinx=lim(x→0)[e^x-e^(-x)]/cosx =0
2
∫dx/[1+x^(1/3)]=∫3x^(2/3)d(x^(1/3)/[1+x^(1/3)]=3∫x^(1/3)d(x^1/3)-3∫d(x^(1/3)+3∫d(x^1/3)/(1+x^1/3)=(3/2)x^(2/3)-3x^(1/3)+3ln(1+x^(1/3)+C
3
f(2x+1)=xe^x, t=2x+1,x=(t-1)/2
f(t)=[(t-1)/2]e^[(t-1)/2]

∫f(t)dt=∫[(t-1)/2]e^[(t-1)/2]dt=∫(t-1)d[e^(t-1)/2]=(t-1)*e^[(t-1)/2] -∫e^[(t-1)/2]d(t-1)=[(t-1)-2]e^[(t-1)/2]+C=(t+1)e^[(t-1)/2]+C
4
∫dx/√x=2√x+C
∫[1,+∞]dx/√x=∞
5
x=sint,dx=cost dt
∫dx/√(1-x^2)=∫costdt/cost=t=arcsinx
∫[-1,1]dx/√(1-x^2)=π/2-(-π/2)=π
6
(1+n)/(1+n^2)
n>1时
1/n<(1+n)/(1+n^2)<1/(n-1)
lim(n→∞)1/n=0 ; lim(n→∞)1/(n-1)=0==>lim(n→∞)(1+n)/(1+n^2)=0
因为an=(1+n)/(1+n^2)上下有界,所以级数收敛
7
收敛域-3≤x≤3, |x^(n-1)/[3^(n-1)]|≤1
lim(n→∞)x^(n-1)/[n*3^(n-1)]=0
8
-1/2≤x≤1/2时
n(n+1)x^n收敛
9.(1-x^n)/(1-x)=1+x+x^2+...+x^(n-1);
收敛域 |x|≤1,
lim(n→∞)(1-x^n)/(1-x)=1/(1-x)=lim(n→∞)1+x+x^2+...+x^(n-1)=∑ (i=1,∞) x^(i-1)
1/(3-x)=(1/3)*[1/(1-x/3)]
=(1/3)*∑ (i=1,∞) (x/3)^(i-1)

Answer 2

1
∫[e^t+e^(-t)]dt=e^t-e^(-t)
lim(x→0) ∫[0,x] [e^t+e^(-t)]dt /cosx=lim(x→0) [e^x-e^(-x)]/cosx =0
2
∫dx/[1+x^(1/3)]=∫3x^(2/3)d(x^(1/3)/[1+x^(1/3)]=3∫x^(1/3)d(x^1/3)-3∫d(x^(1/3)+3∫d(x^1/3)/(1+x^1/3)=(3/2)x^(2/3)-3x^(1/3)+3ln(1+x^(1/3)+C
3
f(2x+1)=xe^x, t=2x+1,x=(t-1)/2
f(t)=[(t-1)/2]e^[(t-1)/2]

∫f(t)dt=∫[(t-1)/2]e^[(t-1)/2]dt=∫(t-1)d[e^(t-1)/2]=(t-1)*e^[(t-1)/2] -∫e^[(t-1)/2]d(t-1)=[(t-1)-2]e^[(t-1)/2]+C=(t+1)e^[(t-1)/2]+C
4
∫dx/√x=2√x+C
∫[1,+∞]dx/√x=∞
5
∫dx/√(1-x^2)=arcsinx
∫[-1,1]dx/√(1-x^2)=π/2-(-π/2)=π
6
(1+n)/(1+n^2)
n>1时
1/n<(1+n)/(1+n^2)<1/(n-1)
lim(n→∞)1/n=0 lim(n→∞)1/(n-1)=0
因为an=(1+n)/(1+n^2)上下有界,所以级数收敛
7
收敛域-3≤x≤3
lim(n→∞)x^(n-1)/[n*3^(n-1)]=0
8
-1/2≤x≤1/2时
n(n+1)x^n收敛 回答者： drug2009 | 八级 | 2011-5-6 16:47 | 检举

1
∫[e^t+e^(-t)-2]dt=e^x-e^(-x)-2x
lim(x→0) ∫[0,x] [e^t+e^(-t)]dt /(1-cosx)=lim(x→0) [e^x-e^(-x)-2x]/(1-cosx)
=lim(x→0) [e^x+e^(-x)-2]/sinx=lim(x→0)[e^x-e^(-x)]/cosx =0
2
∫dx/[1+x^(1/3)]=∫3x^(2/3)d(x^(1/3)/[1+x^(1/3)]=3∫x^(1/3)d(x^1/3)-3∫d(x^(1/3)+3∫d(x^1/3)/(1+x^1/3)=(3/2)x^(2/3)-3x^(1/3)+3ln(1+x^(1/3)+C
3
f(2x+1)=xe^x, t=2x+1,x=(t-1)/2
f(t)=[(t-1)/2]e^[(t-1)/2]

∫f(t)dt=∫[(t-1)/2]e^[(t-1)/2]dt=∫(t-1)d[e^(t-1)/2]=(t-1)*e^[(t-1)/2] -∫e^[(t-1)/2]d(t-1)=[(t-1)-2]e^[(t-1)/2]+C=(t+1)e^[(t-1)/2]+C
4
∫dx/√x=2√x+C
∫[1,+∞]dx/√x=∞
5
x=sint,dx=cost dt
∫dx/√(1-x^2)=∫costdt/cost=t=arcsinx
∫[-1,1]dx/√(1-x^2)=π/2-(-π/2)=π
6
(1+n)/(1+n^2)
n>1时
1/n<(1+n)/(1+n^2)<1/(n-1)
lim(n→∞)1/n=0 ; lim(n→∞)1/(n-1)=0==>lim(n→∞)(1+n)/(1+n^2)=0
因为an=(1+n)/(1+n^2)上下有界,所以级数收敛
7
收敛域-3≤x≤3, |x^(n-1)/[3^(n-1)]|≤1
lim(n→∞)x^(n-1)/[n*3^(n-1)]=0
8
-1/2≤x≤1/2
n(n+1)x^n
9.(1-x^n)/(1-x)=1+x+x^2+...+x^(n-1);
|x|≤1,
lim(n→∞)(1-x^n)/(1-x)=1/(1-x)=lim(n→∞)1+x+x^2+...+x^(n-1)=∑ (i=1,∞) x^(i-1)
1/(3-x)=(1/3)*[1/(1-x/3)]
=(1/3)*∑ (i=1,∞) (x/3)^(i-1)

就是

Answer 3

1
∫[e^t+e^(-t)]dt=e^t-e^(-t)
lim(x→0) ∫[0,x] [e^t+e^(-t)]dt /cosx=lim(x→0) [e^x-e^(-x)]/cosx =0
2
∫dx/[1+x^(1/3)]=∫3x^(2/3)d(x^(1/3)/[1+x^(1/3)]=3∫x^(1/3)d(x^1/3)-3∫d(x^(1/3)+3∫d(x^1/3)/(1+x^1/3)=(3/2)x^(2/3)-3x^(1/3)+3ln(1+x^(1/3)+C
3
f(2x+1)=xe^x, t=2x+1,x=(t-1)/2
f(t)=[(t-1)/2]e^[(t-1)/2]

∫f(t)dt=∫[(t-1)/2]e^[(t-1)/2]dt=∫(t-1)d[e^(t-1)/2]=(t-1)*e^[(t-1)/2] -∫e^[(t-1)/2]d(t-1)=[(t-1)-2]e^[(t-1)/2]+C=(t+1)e^[(t-1)/2]+C
4
∫dx/√x=2√x+C
∫[1,+∞]dx/√x=∞
5
∫dx/√(1-x^2)=arcsinx
∫[-1,1]dx/√(1-x^2)=π/2-(-π/2)=π
6
(1+n)/(1+n^2)
n>1时
1/n<(1+n)/(1+n^2)<1/(n-1)
lim(n→∞)1/n=0 lim(n→∞)1/(n-1)=0
因为an=(1+n)/(1+n^2)上下有界,所以级数收敛
7
收敛域-3≤x≤3
lim(n→∞)x^(n-1)/[n*3^(n-1)]=0
8
-1/2≤x≤1/2时
n(n+1)x^n收敛本回答被提问者采纳