Question

大学文科数学，求不定积分

Answer 1

换元,令√[(a+x)/(a-x)]=t, (a+x)/(a-x)=t^2, a+x=t^2(a-x), a+x=at^2-xt^2, x+xt^2=at^2-a, x(1+t^2)=a(t^2-1)
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