Question

求如图所示的不定积分，在线等，多谢！

Answer 1

换元,令√[(a+x)/(a-x)]=t,则x=a(t^2-1)/(t^2+1),dx=4at/(t^2+1)^2 dt
原积分
= ∫ t*4at/(t^2+1)^2 dt
=4a ∫ t^2/(t^2+1)^2 dt
=4a [∫1/(t^2+1) dt -∫1/(t^2+1)^2dt]
再换元,令t=tanu,u=arctant,dt=1/(cosu)^2.sinu=t/√(1+t^2),cosu=1/√(1+t^2).则上式
=4a [arctant - ∫ (cosu)^2 du]
=4a [arctant - ∫ (1+cos2u)/2 du]
=4a [arctant - u/2-sin2u/4 +C]
=2a [2arctant - u-sinucosu +C]
=2a [2arctant - arctant-t/(1+t^2) +C]
=2aarctan√[(a+x)/(a-x)]-√(a^2-x^2) + C

追问

多谢！