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

(1)xd(x)/(x+1)(x+2)(x+3)

(2)dx/3+sin^2x
(3)dx/1+[3倍根号x+1]
(4)[(根号x)^3]-1/(根号x+1)*dx
(5)(根号x+1)-1/(根号x+1)+1*dx
(6)dx/根号x+4倍根号x
前面都有积分符号，详细过程。谢谢

Answer 1

(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