Question

1/（x²-2x+5）的不定积分

Answer 1

1/[(x-1)^2+4]
设x-1＝2tant
原式＝1/2∫1/(1+t^2)dt＝1/2arctant+C＝1/2arctan[(x-1)/2]+C

Answer 2

∫ 1/（x²-2x+5）dx
=∫ 1/[(x-1)²+2²] dx
=1/4∫1/[(x/2-1/2)²+1] dx
=1/2∫1/[(x/2-1/2)²+1] d(x/2-1/2)
=1/2 arc tan(x/2-1/2)+C