关于不定积分2∫ln[x+(x^2+1)^1/2]dx;做不出来,

Question

关于不定积分2
∫ln[x+(x^2+1)^1/2]dx;做不出来,

Answer 1

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