Question

一个不定积分计算题，求解

(y2 + 2xy - x2) / (x2 + y2)2 对x求不定积分
要求是不用换元法计算，答案上面也没说过程
具体过程写一下

Answer 1

没可能不用换元法计算吧？
∫(y²+2xy-x²)/(x²+y²)² dx
令x=ytanθ,dx=ysec²θdθ
(x²+y²)²=y⁴sec⁴θ
∫(-x²+2xy+y²)/(x²+y²)² dx
=(1/y)∫(-tan²θ+2tanθ+1)/(sec²θ) dθ
=(1/y)∫(sin2θ+cos2θ)dθ
=(1/2y)(sin2θ-cos2θ)+C
=(1/2y)[(x²+2xy-y²)/(x²+y²)]+C
=(x-y)/(x²+y²)+1/2y+C
=(x-y)/(x²+y²)+C'
这个是分部积分法，有点硬来的，其实原函数是透过求导的商法则得到被积函数：
∫(-x²+2xy+y²)/(x²+y²)² dx
=∫[(-2x³+3x²y+y³)+(x-y)(x²+y²)]/[x(x²+y²)²] dx
=∫x*(-2x³+3x²y+y³)/[x(x²+y²)²] dx+∫(x-y)/[x(x²+y²)] dx
=∫x d{(x-y)/[x(x²+y²)]}+∫(x-y)/[x(x²+y²)] dx
=x*(x-y)/[x(x²+y²)]-∫(x-y)/[x(x²+y²)]+∫(x-y)/[x(x²+y²)] dx
=(x-y)/(x²+y²)+C

Answer 2

我才初三！！！