求一道不定积分的题……看似不是很难.可是就是做不来=.=∫ln(x^2+2)dx不要跳步.

右边通分=[a(x-2)+b(x-1)]/(x-1)(x-2)
=[(a+b)x-(2a+b)]/(x-1)(x-2)
所以3x-4=(a+b)x-(2a+b)
a+b=3
2a+b=4
所以a=1,b=2

原式=xln(x²+2)-∫xdln(x²+2)=xln(x²+2)-∫2x²/(x²+2)dx=xln(x²+2)-2∫[1-2/(x²+2)]dx=xln(x²+2)-2x+4∫dx/(x²+2)令x=√2adx=√2da则∫dx/(x²+2)=√2∫da/(2a&su...