∫( 1+2x+3x^2 +4x^3 +.........) dx , (0<∣x∣<1)


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Question Number 41651 by rahul 19 last updated on 10/Aug/18

$\int (1 + 2 x + 3 x^{2} + 4 x^{3} + . . . . . . . . .) d x,$ $(0 <∣ x ∣< 1)$
Commented bymaxmathsup by imad last updated on 10/Aug/18

$\int (1 + 2 x + 3 x^{2} + 4 x^{3} + . . .) d x = \int (\sum_{n = 1}^{\infty} {n x}^{n - 1}) d x$ $= \sum_{n = 1}^{\infty} n \int x^{n - 1} d x = \sum_{n = 1}^{\infty} x^{n} = \frac{1}{1 - x} - 1 + c = \frac{x}{x - 1} + c w i t h ∣ x ∣< 1 .$
	Answered by alex041103 last updated on 10/Aug/18

	$1 + 2 x + 3 x^{2} + . . . = \underset{n = 0}{\sum^{\infty}} (n + 1) x^{n}$ $\Rightarrow \int (1 + 2 x + 3 x^{2} + 4 x^{3} + . . . . . . . . .) d x =$ $= \int \underset{n = 0}{\sum^{\infty}} (n + 1) x^{n} d x = \underset{n = 0}{\sum^{\infty}} \int (n + 1) x^{n} d x =$ $= C + \underset{n = 0}{\sum^{\infty}} \frac{n + 1}{n + 1} x^{n + 1} = C + \underset{n = 0}{\sum^{\infty}} x^{n + 1} =$ $= C + (\underset{n = 0}{\sum^{\infty}} x^{n}) - 1 = (\underset{n = 0}{\sum^{\infty}} x^{n}) + C_{1}$ $F o r ∣ x ∣< 1 : \underset{n = 0}{\sum^{\infty}} x^{n} = \frac{1}{1 - x}$ $\Rightarrow \int (1 + 2 x + 3 x^{2} + 4 x^{3} + . . . . . . . . .) d x = \frac{1}{1 - x} + C$
	Commented byrahul 19 last updated on 10/Aug/18

	$thanks sir!$
	Answered by rahul 19 last updated on 10/Aug/18

	$\int \frac{d}{d x} (x + x^{2} + x^{3} + . . . . . . . . .) d x$ $\Rightarrow x + x^{2} + x^{3} + . . . . . . . . . .$ $\Rightarrow \frac{x}{1 - x} .$ $W {hat}^{'} s wrong in this ?$
	Commented byalex041103 last updated on 10/Aug/18

	$N o t h i n g i s w r o n g . I f y o u l o o k m o r e$ $c a r e f u l l y y o i w i l l s e e t h a t :$ $\frac{x}{1 - x} = - \frac{1 - x - 1}{1 - x} = - (\frac{1 - x}{1 - x} - \frac{1}{1 - x}) =$ $= \frac{1}{1 + x} - 1 = \frac{1}{1 + x} + C$ $A s y o u c a n s e e t h e i n t e g r a t i o n c o n s t a n t$ $i s s t h v e r y i m o r t a n t$
	Commented byrahul 19 last updated on 10/Aug/18

	$ohh, yes u r right sir!$
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