急！求不定积分 高数作业，要详细过程！好的有加分O(∩_∩)O谢谢

Question

1. ∫x^2/9+x^2 dx
2. ∫x^2+1/(x+1)^2*(x-1) dx
3. ∫x/x^2+2x+2 dx
4. ∫dx/x*根号下2x+1
5. ∫(根号下x+1)-1/(根号下x+1)+1 dx
6. ∫1/根号下1+e^x dx

Answer 1

1. ∫x^2dx/(9+x^2)=∫[1-9/(9+x^2)]dx=x-3∫dx/3/(1+(x/3)^2)=x-3arctan(x/3)+C

2. ∫(x^2+1)dx/[(x+1)^2*(x-1) ]=∫[(x+1)^2-2(x-1)-2]dx/[(x+1)^2(x-1)]
=∫dx/(x-1)-2∫dx/(x+1)^2-∫[(x+1)-(x-1)]dx/[(x+1)^2(x-1)]
=ln|x-1|+2/(x+1)-(1/2)∫[(x+1)-x-1)]dx/[(x+1)(x-1)] +∫dx(x+1)^2
=ln|x-1|+2/(1+x)-(1/2)ln|x-1|+(1/2)ln|x+1|-1/(1+x)+C
=(1/2)ln|x-1|+1/(1+x)+(1/2)ln|x+1|+C

3. ∫xdx/x^2+2x+2 =(1/2)∫(2x+2)dx/(x^2+2x+2) -∫dx/[(x+1)^2+1]
=(1/2)ln|x^2+2x+2|-arctan(x+1)+C

4. ∫dx/[x*√(2x+1)]=(1/2)∫xd(2x+1)/√(2x+1)=∫xd√(2x+1)=x√(2x+1)-∫(2x+1)dx
=x√(2x+1)-(1/2)*(2/3)√(2x+1)^3+C

5. ∫[(√(x+1)-1/(√x+1)+1 ]dx=∫[√x+1-1]^2dx/[√(x+1)^2-1] =∫[x+2-2√(x+1)]dx
=x^2/2+2x-2*(2/3)√(x+1)^3+C

6. ∫dx/∫√(1+e^x )=∫e^(-x/2)dx/√[1+e^(-x)]=-2∫de^(-x/2)/√{1+[e^(-x/2)]^2}
=-2arcsin [e^(-x/2)] +C

追问

很感谢你的回答，不过这些题目都是有理函数的不定积分，有2道题你的答案和书上的答案不一样额··· 5题是x-4根号下x+1+4ln[(根号下x+1)+1]+C 6题是ln (根号下1+e^x)-1/(根号下1+e^x)+1+C 还有第1题我打错了，上面是x^3 可以请你再看一下这3道题吗？

追答

谢谢提醒，发现有误算，修改如下：
1
∫x^3dx/(9+x^2)
=(1/2)∫x^2dx^2/(9+x^2)
=(1/2)∫(x^2+9)-9dx^2/(9+x^2)
=(1/2)x^2-(9/2)ln(x^2+9)+C

2 ∫(x^2+1)dx/[(x+1)^2*(x-1) ]=∫[(x+1)^2-2(x-1)-2]dx/[(x+1)^2(x-1)]
=∫dx/(x-1)-2∫dx/(x+1)^2-∫[(x+1)-(x-1)]dx/[(x+1)^2(x-1)]
=ln|x-1|+2/(x+1)-(1/2)∫[(x+1)-(x-1)]dx/[(x+1)(x-1)] +∫dx(x+1)^2
=ln|x-1|+2/(1+x)-(1/2)ln|x-1|+(1/2)ln|x+1|-1/(1+x)+C
=(1/2)ln|x-1|+1/(1+x)+(1/2)ln|x+1|+C

5∫[(√(x+1)-1]dx/(√x+1)+1]
=∫[√(x+1)-1]d[√(x+1)^2]/[√(x+1)+1]
=∫2√(x+1)*[√(x+1)-1]d√(x+1)) /[√(x+1)+1]
=∫2√(x+1)^2+2√(x+1) -4√(x+1)-4 +4] d(√(x+1)/ [√(x+1)+1]
=∫2√(x+1)d√(x+1) -4√(x+1)+4∫d√(x+1)/[√(x+1)+1]
=x+1-4√(x+1)+4ln|√(x+1)+1| +C

3
∫dx/√(1+e^x)=∫e^(-x/2)dx/√(1+e^(-x))=-2∫de^(-x/2)/√[1+(e^(-x/2))^2]
e^(-x/2)=tanu de^(-x/2)=secu^2du =-2∫secu^2du/secu
=-2∫du/cosu
=2∫dsinu/[(1-sinu)(1+sinu)]
=∫dsinu/(1-sinu)+∫dsinu/(1+sinu)
=ln|(1+sinu)/(1-sinu)| +c
=ln|(1+sinu)^2/cosu| +c
=2ln|1+sinu/cosu| +C
=2ln|secu+tanu|+c
=2ln|√(1+e^(-x)) +e^(-x/2)|+C