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