Let f be a function defined in the interval (1,\infty) as f(x)=\integral_{e} ^{x^2} t/ln(t) dt. Find the equation of the tangent line to the graph of f at the point whose x-coordinate is sqrt{e}.

The equation of the tangent line to the graph of the function y=f(x) at the point whose x-coordinate is sqrt{e} is given by y-y_f=m(x-x_f), where x_f=sqrt{e}, y_f=f(sqrt{e})=integral_{e}^{e} t/ln(t) dt = 0 and m=f'(sqrt{e})=[x^22x/ln(x^2)]_x=sqrt{e}=2esqrt{e}/ln(e)=2esqrt{e}. To calculate m we used the Fundamental theorem of calculus.Then, the tangent line has equation y=2esqrt{e}(x-sqrt{e}), so y=2e*sqrt{e}x-2e^2.