三角换元脱根号，换元x=tanu，=∫tan²usecudtanu=∫tan²usec³udu=∫tanusec²udsecu=1/3∫tanudsec³u=tanusec³u/3-1/3∫sec³udtanu=tanusec³u/3-1/3∫tan²usec³u+sec³udu=(tanusec³u-∫sec³udu)/4=(2tanusec³u-secutanu-ln|secu+tanu|)/8+C
