$\int (1 + 2 x + 3 x^{2} + 4 x^{3} + . . .) d x =$

(1 + x)^{- 1} + c

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(1 - x)^{- 1} + c

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(1 - x)^{- 1} - 1 + c

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None of these

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Solution

The correct option is B $(1 - x)^{- 1} + c$
$\int (1 + 2 x + 3 x^{2} + 4 x^{3} + . .) d x$
$= \int (1 - x)^{2} d x [(1 - x)^{2} = 1 + 2 x + 3 x^{2}]$
$= \int \frac{d x}{{(1 - x)}^{2}}$
$= (1 - x)^{- 1} + c$
$∴ \int (1 + 2 x + 3 x^{2} + . . . . .) d x = (1 - x)^{- 1} + c$

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