Evaluating function with integral

$\begin{aligned} & xf(x) = x+ \displaystyle \int_1^x f(t) \,dt \\ \implies & xf'(x) + f(x) = 1 + f(x) & \small\color{#3D99F6}\text{Differentiating both sides wrt x} \\ \implies & f'(x) = \dfrac{1}{x} \\ \implies & f(x) = \ln x + C \\ \implies & f(x) = \ln x + 1 & \small\color{#3D99F6}\text{Putting } x=1 \text{ in the original equation we get } f(1) =1 \text{ hence C=1} \\ \implies & f(e^k) = k+1 \\ \implies & \displaystyle \sum_{k=1}^{10} f(e^k) = \displaystyle \sum_{k=1}^{10} (k+1) = \dfrac{10 \times 11}{2} + 10 = \boxed{65} \end{aligned}$

at x=1 f(1)=1 by applying $\frac{d}{dx}$ on both side (we have to apply labnize) $f(x)+x \times f^|(x) = 1+f(x)$ = $f^|(x)=\frac{1}{x}$ by integrating we get = $f(x)=log(x)+c$ by applying f(1)=1 we get c=1 therefore f(x)=log(x)+1 $\sum_{k=1}^10 f(e^{k}=\sum_{k=1}^10 k +\sum_{k=1}^10 1$ = 55+10 =65