The equation of the tangent to the curve $y=\int_{x^2}^{x^3}\frac{dt}{1+t^2}$ at x = 1 is:

$\sqrt{2}y+1=x$ $\sqrt{3}x+1=y$ $\sqrt{3}x+1+\sqrt{3}=y$ None of these

$\sqrt{2}y+1=x$

$\frac{dy}{dx}|_{x=1}=\frac{3x^2}{\sqrt{1+x^6}}-\frac{2x}{\sqrt{1+x^4}}|_{x=1}=\frac{1}{\sqrt{2}}$ Equation of tangent $\sqrt{2}y+1=x$