用分部积分法求ln【x+√（x²+1)】dx

网友回答

令x=tant,则
原式=∫ln（tant+sect)dtant
=tant*In(tant+sect)-∫tantsectd
=tant*In(tant+sect)-∫dsect
=tant*In(tant+sect)-sect
=x*ln(x+√(1＋x²))-√(1＋x²)+C.
======以下答案可供参考======
供参考答案1:
原式=xln(x+√(x^2+1))-∫x*1/(x+√(x^2+1))*(1+1/2*1/√(x^2+1)*2x)dx=xln(x+√(x^2+1))-∫x/√(x^2+1)dx=xln(x+√(x^2+1))-1/2∫d(x^2+1)/√(x^2+1)=xln(x+√(x^2+1))-√(x^2+1)+C