$\displaystyle F(x) = \int_1^{\ln ({x^2})} {\frac{1}

{t}dt}

$

generalized theorem: $\displaystyle F(x) = \int_{h(x)}^{g(x)} {f(t)dt}

$$\displaystyle F'(x) = f(g(x))*g'(x) = f(h(x))*h'(x)

$

I tired:

$\displaystyle \frac{1}

{{\ln ({x^2})}}*\frac{{2x}}

{{{x^2}}} = \frac{{2x}}

{{{x^2}\ln ({x^2})}}

$

$\displaystyle F'(e) = \frac{{2e}}

{{{e^2}\ln {{(e)}^2}}} = \frac{{2e}}

{{2{e^2}}} = .368

$

is this correct?