<s>[INST] Values of $6 + \int_a^x \frac{f(t)}{t^2} dt = 2 \sqrt{x}$ Let $f$ and $a$ such that $6 + \int_a^x \frac{f(t)}{t^2} dt = 2 \sqrt{x}$.
I need to find the values of $f$ and $a$ that satisfies this condition.
For this i tried:
$F(x) = 6 + \int_a^x \frac{f(t)}{t^2} dt = 2 \sqrt{x}$ then $F´(x) = \frac{f(x)}{x^2} = \frac{1}{\sqrt{x}}$ since $ \frac{d}{dx} \int_a^x \frac{f(t)}{t^2} dt = \frac{f(x)}{x^2}$ and here i stuck , how can i conclude in get de values of $f$ and $a$  please some healp, thanxs for your time.
 [/INST] Put $x=a$ in $6 + \int_a^x \frac{f(t)}{t^2} dt = 2 \sqrt{x}$. This gives you that
$2\sqrt a=6$ and so $a=9$. Differentiating $6 + \int_a^x \frac{f(t)}{t^2} dt = 2 \sqrt{x}$ we get $\frac{f(x)}{x^2}=\frac{1}{\sqrt x}$ which implies that $f(x)=x^{3/2}$. Note that $\int_a^a \frac{f(t)}{t^2} dt=0$.   
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