Question

这个不定积分怎么算？35题

Answer 1

x/(x^2-x-2)dx
x/(x-2)(x+1)
x(1/(x-2)(x+1）
=1/3xx*(1/(x-2)-1/(x+1））
=1/3(x/(x-2)-x/(x+1)dx
x/(x-2)dx
=x/(x-2)d(x-2)
=xdln(x-2)
=xln(x-2)-ln(x-2)dx
=xln(x-2)-ln(x-2)d(x-2)
=xln(x-2)-((x-2)(ln(x-2)-1)
=xln(x-2)-(x-2)ln(x-2)+x-2
=ln(x-2)(x-x+2)+x-2
=2ln（x-2)+x-2
x/(x+1）dx
=x/(x+1)d(x+1)
=xdln(x+1)
=xln(x+1)-ln(x+1)dx
=xln(x+1)-ln(x+1）d(x+1）
=xln(x+1)-[(x+1）（ln(x+1)-1)
=xln(x+1）-(x+1)ln(x+1)+x+1
=-ln(x+1）+x+1
原是=1/3[2ln（x-2)+x-2+ln(x+1)-x-1]+C
=1/3(2ln（x-2)+ln(x+1)-3]+C
=2/3ln（x-2)+1/3ln(x+1)-1+C
=2/3ln（x-2)+1/3ln(x+1)+C

Answer 2

x/(x^2-x-2)=x/[(x+1)(x-2)]
=(1/3)[1/(x+1)+2/(x-2)]
原式=(ln|x+1|+2ln|x-2|)/3+C

Answer 3

这个😣