求不定积分∫√(1+x²)dx令x=tanu,则dx=sec²udu,于是原式=∫sec³udu=∫secud(tanu)=secutanu-∫tanud(secu)=secutanu-∫tan²usecudu=secutanu-∫(sec²u-1)secudu=secutanu-∫sec³udu+...