Let $f(x)=\int_{0}^{x} e^{t} f(t) d t+e^{x}$ be a differentiable function for all $x \in R$. Then $\mathrm{f}(\mathrm{x})$ equals.
Correct Option: 1
Given, $f(x)=\int_{0}^{x} e^{t} f(t) d t+e^{x} \ldots(1)$
Differentiating both sides w.r.t $x f^{\prime}(x)=e^{x} \cdot f(x)+e^{x}$
(Using Newton Leibnitz Theorem) $\Rightarrow \frac{f^{\prime}(z)}{f(x)+1}=e^{x}$
Integrating w.r.t $x \int \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})+1} \mathrm{dx}=\int \mathrm{e}^{\mathrm{X}} \mathrm{dx}$
$\Rightarrow \ell n(f(x)+1)=e^{x}+c$
Put $x=0$
$2=1+c \quad(\because \mathrm{f}(0)=1$, from equation $(1))$
$\therefore \ln (f(x)+1)=e^{x}+\ln 2-1$
$\Rightarrow f(x)+1=2 \cdot e^{e^{x}-1}$
$\Rightarrow f(x)=2 e^{e^{x}-1}-1$
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