Question

求不定积分(x^2-1)^(1/2) /x^2 dx

Answer 1

楼上怎么起了个头而没有尾啦？
令x=secN，dx=secNtanNdN
cosN=1/x，sinN=√(1-1/x²)=√(x²-1)/x
√(x²-1)=√(sec²N-1)=tanN
∴∫√(x²-1)/x² dx
=∫tanN/sec²N*secNtanN dN
=∫tan²N/secN dN
=∫sin²N/cosN dN
=∫(1-cos²N)/cosN dN
=∫(secN-cosN) dN
=ln|secN+tanN|-sinN+C，代入上面所设的换元
=ln|x+√(x²-1)|-√(x²-1)/x+C

Answer 2

令x=sect
原式=∫tant/(sect)^2d(sect)=∫tant^2/sectdt
=∫sectdt-∫sect^(-1)dt
=∫sectdt-∫costdt
= ln│sect+tant│+sint+C