int (1 + 2x + 3x^2 + 4x^3 + dots) , dx = | vedclass

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Question

(B) The given series is $1 + 2x + 3x^2 + 4x^3 + \dots = \sum_{n=1}^{\infty} n x^{n-1}$.This is the derivative of the geometric series $\sum_{n=0}^{\in

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The given series is $1 + 2x + 3x^2 + 4x^3 + \dots = \sum_{n=1}^{\infty} n x^{n-1}$.This is the derivative of the geometric series $\sum_{n=0}^{\infty} x^n = (1 - x)^{-1}$ for $|x| Thus,$1 + 2x + 3x^2 + 4x^3 + \dots = \frac{d}{dx} (1 - x)^{-1} = (1 - x)^{-2}$.Now,we integrate this expression:$\int (1 - x)^{-2} \, dx = \frac{(1 - x)^{-1}}{-1 \cdot (-1)} + c = (1 - x)^{-1} + c$.

$\int (1 + 2x + 3x^2 + 4x^3 + \dots) \, dx = $

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