Question

关于有理函数的不定积分的题目，怎样知道用什么方法解题

Answer 1

有理函数积分主要是部分分式的分解：
设q(x)=c(x-a)^α...(x-b)^β(x^2+px+q)^λ...(x^2+rx+s)^μ
(其中p^2-4q<0,...,r^2-4s<0.).
那么真分式p(x)/q(x)可以分解成如下部分分式之和：
p(x)/q(x)=a1/(x-a)^α+a2/(x-a)^(α-1)+...+a[α]/(x-a)+...+
+b1/(x-b)^β+b2/(x-b)^(β-1)+...+b[β]/(x-b)+
(m1x+n1)/(x^2+px+q)^λ+...+(m[λ]x+n[λ])/(x^2+px+q)+......+
(r1x+s1)/(x^2+rx+s)^μ+...+(r[μ]x+s[μ])/(x^2+rx+s).
x/[(x+1)(x+2)(x+3)]=a/(x+1)+b/(x+2)+c/(x+3),
x=a(x+2)(x+3)+b(x+1)(x+3)+c(x+1)(x+2).
令x=-1,得a=-1/2,
令x=-2,得b=2,
令x=-3,得c=-3/2,
x/[(x+1)(x+2)(x+3)]=(-1/2)*1/(x+1)+2/(x+2)-(3/2)*1/(x+3),
或由x=(a+b+c)x^2+(5a+4b+3c)x+(6a+3b+2c),
比较系数得a+b+c=0,5a+4b+3c=1,6a+3b+2c=0,
解出a,b,c.
3/(x^3+1)=1/(x+1)(x^2-x+1)=a/(x+1)+(mx+n)/(x^2-x+1),
3=a(x^2-x+1)+(mx+n)(x+1).
令x=-1,得a=1,
(mx+n)(x+1)=3-a(x^2-x+1)=-x^2+x-2=-(x-2)(x+1),
mx+n=-x+2,m=-1,n=2.
3/(x^3+1)=1/(x+1)-(x-2)/(x^2-x+1).

Answer 2

(1)把被积函数分解：x/{(x+1)(x+2)(x+3)}=(x+1-1)/{(x+1)(x+2)(x+3)}=1/{(x+2)(x+3)}-1/{(x+1)(x+2)(x+3)}
=1/(x+2)-1/(x+3)-1/(x+1){1/(x+2)-1/(x+3)}=1/(x+2)
-
1/(x+3)
-
1/(x+1)
+
1/(x+2)
+
1/2{1/(x+1)
-
1/(x+3)}
=2/(x+2)-1.5/(x+3)-0.5/(x+1),积分后结果=
2
ln(x+2)-1.5ln(x+3)-0.5ln(x+1)+C
(2)dx/{3+sin^2(x)}={sin^2(x)+cos^2(x)}/{4sin^2(x)+3cos^2(x)}
dx={tan^2(x)+1}/{4tan^2(x)+3}
dx，设t=tan(x)
则x=arctan(t),dx=1/(1+t^2)
dt，所以上式=（t^2+1）/(4t^2+3)*1/(t^2+1)
dt=1/(4t^2+3)
dt
=1/{2sqrt(3)}
1/{(t/(sqrt(3)/2))^2+1}d(t/(sqrt(3)/2))积分后=1/{2sqrt(3)}*arctan{t/(sqrt(3)/2},用t=tan(x)喊回来
得到
1/{2sqrt(3)}
arctan{2sqrt(3)tan(x)/3}+C
(3)原式=dx/{1+3sqrt(x+1)}=d{sqrt(x+1)}^2/{1+3sqrt(x+1)}，设t=sqrt(x+1)则原式=2tdt/(3t+1)=2/3
{1-1/(3t+1)}dt积分后=2/3*t-2/9*ln(3t+1)+C=2/3*sqrt(x+1)-2/9*ln(3sqrt(x+1)+1)+C
(4)打字太累了，用t=sqrt(x)替换，只需要做多项式的积分就可以了
（5）设t=sqrt(x+1),做多项式积分
（6）令t=sqrt(x+4)做多项式积分
（4）（5）（6）都是变换后展开做简单多项式不定积分