Question

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
过程请详细。。不要跳步。。。。。。。。。。。。。

Answer 1

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=√2a
dx=√2da
则∫dx/(x²+2)
=√2∫da/(2a²+2)
=√2/2∫da/(a²+1)
=√2/2*arctan(a²+1)
=√2/2*arctan(x²/2+1)

所以原式=xln(x²+2)-2x+2√2arctan(x²/2+1)+C

Answer 2

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