用换元法求不定积分∫(1/根号下4+x^2)dx

Question

用换元法求不定积分
∫(1/根号下4+x^2)dx

Answer 1

∫1/(4+x^2)^(1/2)dx
=∫1/(1+(x/2)^(1/2)d(x/2) t=x/2
=∫1/(1+t^2)^(1/2)dt
=∫(1/1+(tana)^2)^(1/2)d(tana)
=∫cosa(1+tanatana)da
=∫(1/cosa)da
=2∫1/[1-tan(a/2)^2]d(tana/2)
=ln(tan(a/2)+1)-ln(tan(a/2)-1)+C
=ln(x+(x^2+4))+C