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\(\int\left(1+\frac{x}{1 !}+\frac{x^2}{2 !}+\ldots \infty\right) d x=\)
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Verified Answer
The correct answer is:
\(e^x+c\)
\(\begin{gathered}
\int\left(1+\frac{x}{1 !}+\frac{x^2}{2 !}+\ldots \ldots\right) d x \\
=\int e^x \cdot d x=\left(e^x+c\right)
\end{gathered}\)
\int\left(1+\frac{x}{1 !}+\frac{x^2}{2 !}+\ldots \ldots\right) d x \\
=\int e^x \cdot d x=\left(e^x+c\right)
\end{gathered}\)
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