Let u=x+√(1+x2), then du/dx=1+x/√(1+x2)=(x+√(1+x2))/√(1+x2)=u(1+x2)-½.
Let v=ln(u), dv/du=1/u; y=v2,
y(1)=(dy/dv)(dv/du)(du/dx)=(2v)(1/u)(u(1+x2)-½)=2v(1+x2)-½
dv/dx=(dv/du)(du/dx)=(1+x2)-½.
y(2)=d2y/dx2=-2vx(1+x2)-3/2+(1+x2)-½(2dv/dx)=-2vx(1+x2)-3/2+2(1+x2)-1,
y(2)=2(1+x2)-1(1-vx(1+x2)-½).
y(3)=2(1+x2)-1(-(1+x2)-½(v+x(1+x2)-1)+vx2(1+x2)-3/2),
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