The equation of the tangent to the curve $y=\int_{x^2}^{x^3}\frac{dt}{1+t^2}$ at x = 1 is:
Answer & explanation
Correct answer: option 1
$\frac{dy}{dx}|_{x=1}=\frac{3x^2}{\sqrt{1+x^6}}-\frac{2x}{\sqrt{1+x^4}}|_{x=1}=\frac{1}{\sqrt{2}}$
Equation of tangent $\sqrt{2}y+1=x$