Question

求不定积分，这道题有哪位大神会

第三题，我滴妈，愁死我了，要不要因式分解？？

Answer 1

∫ (x+1)/(x^2-2x+5) dx
=(1/2)∫ (2x-2)/(x^2-2x+5) dx +2∫ dx/(x^2-2x+5)
=(1/2)ln|x^2-2x+5| +2∫ dx/(x^2-2x+5)
=(1/2)ln|x^2-2x+5| +arctanu[( x-1)/2] + C
consider
x^2-2x+5 =(x-1)^2 +4
let
x-1 = 2tanu
dx= 2(secu)^2 du
∫ dx/(x^2-2x+5)
= (1/2)∫ du
=(1/2) arctanu[( x-1)/2] + C

Answer 2