1/(x^2乘以根号下(1+x^2))的不定积分怎么算 用变量替换法

Answer 1

解：
令x=tanα，dx=sec²α dα
则∫1/[x²√(1+x²)]dx
=∫1/[tan²α√(1+tan²α)] sec²xdα
=∫cos⁵α/sin²α dα
=∫cos⁴α/sin²α d(sinα)
=∫(1-sin²α)²/(sin²) d(sinα)
=∫(sinα-1/sinα)² d(sinα)
=∫(sin²α-2+1/sin²α) d(sinα)
=sin³α/3-2sinα-1/sinα+C
=√[x⁶/(x²+1)³]-2√[x²/(x²+1)]-√[(x²+1)/x²]+C