求不定积分。∫dx/(x^2-x-2) ∫dx/(x^2+x+2) ∫(x+1)dx/(x^2+1) 紧急啊 求帮助啊

Question

如题所述

Answer 1

∫dx/(x^2-x-2) =∫dx/[(x-2)(x+1)] =(1/3)∫dx/(x-2)-(1/3)∫dx/(x+1) =(1/3)ln|x-2|-(1/3)ln|x+1|+C =(1/3)ln|(x-2)/(x+1)|+C ∫dx/(x^2+x+2) =∫dx/[(x+1/2)^2+7/4] =(7√7/8)∫d(2x/√7)/[1+((2/√7)(x+1/2))^2] =(7√7/8)arctan[2(x+1/2)/√7)+C ∫(x+1)dx/(x^2+1) =∫xdx/(x^2+1)+∫dx/(x^2+1) =(1/2)∫d(x^2+1)/(x^2+1)+∫dx/(x^2+1) =(1/2)ln(1+x^2)+arctanx+C

Answer 2

这么简单的化简你都弄不出来？？？骗人的吧∫dx/（x2-x-2）＝1/3∫（1/（x-2）-1/（x+1））dx只提示到这里，下面的题目一样