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+∫secudu
移项得2∫sec³udu=secutanu+∫secudu=secutanu+ln(secu+tanu)+2C
故∫√(1+x²)dx=(1/2)secutanu+(1/2)ln(secu+tanu)+C
=(1/2)x√(1+x²)+(1/2)ln[√(1+x²)+x]+C