Question

不定积分习题：∫（x+sinxcosx）/(cosx-xsinx)^2 dx=

Answer 1

这题相当难的：
假设原函数的分母是cosx - xsinx
则设原函数：y = f(x) / (cosx - xsinx)
dy / dx = [f'(x)(cosx - xsinx) - f(x)(-2sinx - xcosx)] / (cosx - xsinx)²
令f'(x)(cosx - xsinx) + f(x)(2sinx + xcosx) = x + sinxcosx
解了微分方程则得f(x) = xsinx，∴原函数为xsinx / (cosx - xsinx) + C
或者直接积分：
∫ (x + sinxcosx) / (cosx - xsinx)² dx
= ∫ [x(sin²x + cos²x) + sinxcosx] / (cosx - xsinx)² dx
= ∫ (sinxcosx - xsin²x + 2xsin²x + xcos²x) / (cosx - xsinx)² dx
= ∫ [(sinxcosx - xsin²x) + (2xsin²x + xcos²x)] / (cosx - xsinx)² dx
= ∫ sinx / (cosx - xsinx) dx + ∫ x(2sin²x + cos²x) / (cosx - xsinx)² dx
= x*sinx / (cosx - xsinx) - ∫ x d[sinx / (cosx - xsinx)] + ∫ x(2sin²x + cos²x) / (cosx - xsinx)² dx
= xsinx / (cosx - xsinx) - ∫ x(2sin²x + cos²x) / (cosx - xsinx)² dx
+ ∫ x(2sin²x + cos²x) / (cosx - xsinx)² dx，刚好抵消。
= xsinx / (cosx - xsinx) + C

Answer 2

我做到的答案是-1/(1-xtanx)+C

上下同时除以cosx的平方

∫（x/cosx²+tanx）/(1-xtanx)²dx

=∫1/(1-xtanx)²dxtanx

=-1/(1-xtanx)+C