令x=atant, 則a^2+x^2=a^2(1+tant^2)=(asect)^2

dx=d(atant)=a(sect)^2dt.

S√(a^2 ＋x^2)dx= Sasect. a(sect)^2dt =a^2S(sect)^3dt=a^2Ssect d(tant)

=a^2[sect.tant -Stant.(sect)"dt=a^2[sect.tant -Stant.sect.tantdt]

=a^2{.sect.tant -S[(sect)^3-sect]dt}

S√(a^2 ＋x^2)dx= Sasect. a(sect)^2dt=a^2.sect.tant -a^2S[(sect)^3dt + a^2Ssectdt

左右移項合併,

2S(sect)^3dt=.sect.tant + S1/cost dt

S1/cost dt = S1/sin(π/2+t) dt= ln/sect+tant/+C

即S√(a^2 ＋x^2)dx=1/2[sect.tant + ln/sect+tant/]+C

tant=x/a, 則sect=√(a^2+x^2)/a

原函式=1/2{x√(a^2+x^2)/a^2 +ln[(√(a^2+x^2)+x] -lna}+C

原函式=1/2{x√(a^2+x^2)/a^2 +ln[(√(a^2+x^2)+x]}+C