∫ln[x+(1+x^2)^(1/2)]dx(分部积分法怎么求)_数学解答

∫ln[x+(1+x^2)^(1/2)]dx(分部积分法怎么求)

人气：264 ℃　时间：2019-11-16 07:46:11

解答

∫ln[x+(1+x^2)^(1/2)]dx
=xln[x+(1+x^2)^(1/2)]-∫[x(1+x/(1+x^2)^(1/2)]/[x+(1+x^2)^(1/2)]dx
=xln[x+(1+x^2)^(1/2)]-∫[x/(1+x^2)^(1/2)]dx
=xln[x+(1+x^2)^(1/2)]-1/2∫(1+x^2)^(-1/2)d(1+x^2)
=xln[x+(1+x^2)^(1/2)]-(1+x^2)^(1/2)+C
=xln[x+√(1+x^2)]-√(1+x^2)+C