The correct option is A e^x+c We know that e^x=1+x+x^22!+x^33!+...x^nn!+...∞ Hence ∫ (1+x+x^22!+x^33!+...x^nn!+...∞ )dx =∫ e^xdx =e^x+ ...

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Centroid Formula

∫1+x+x2/2!+x3...

Question

$\int (1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots \dots . .) d x =$

e^{x} + c

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- e^{x} + c

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\frac{e^{x}}{2} + c

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- \frac{e^{x}}{2} + c

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Solution

The correct option is A $e^{x} + c$
We know that
$e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . \frac{x^{n}}{n!} + . . . \infty$
Hence
$\int (1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . \frac{x^{n}}{n!} + . . . \infty) d x$
$= \int e^{x} d x$
$= e^{x} + c$

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