Question

一个不定积分计算题,(y2 + 2xy - x2) / (x2 + y2)2 对x求不定积分要求是不用换元法计算,

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