Question

高数不定积分求大侠帮忙 ∫1/[(1+x^2)^(1/2)]dx,求公式推导。结果等于ln[x+(1+x^2)^(1/2)]..

Answer 1

∫1/[(1+x^2)^(1/2)]dx

令x=tant   t=arctanx

∫1/[(1+x^2)^(1/2)]dx

=∫1/[(1+tan^2t)^(1/2)]dtant

=∫sec^2t/sectdt

=∫sectdt

=∫sect(sect+tant)/(sect+tant)dt

=∫(sec^2t+tantsect)/(sect+tant)dt

=∫1/(sect+tant)d(sect+tant)

=ln(sect+tant)+C

=ln(secarctanx+tanarctanx)+C

=ln{[(1+tan^2arctanx)^(1/2)]+x}+C

=ln[(1+x^2)^(1/2)+x]+C