1-2x-3x-2-4x-3-dx-0-lt-x-lt-1-

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

\int (1 + 2 x + 3 x^{2} + 4 x^{3} + \dots \dots \dots) d x,

(0 <∣ x ∣< 1)

Commented by maxmathsup by imad last updated on 10/Aug/18

\int (1 + 2 x + 3 x^{2} + 4 x^{3} + \dots) 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} + \dots = \underset{n = 0}{\sum^{\infty}} (n + 1) x^{n}

\Rightarrow \int (1 + 2 x + 3 x^{2} + 4 x^{3} + \dots \dots \dots) 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} + \dots \dots \dots) d x = \frac{1}{1 - x} + C

Commented by rahul 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} + \dots \dots \dots) d x

\Rightarrow x + x^{2} + x^{3} + \dots \dots \dots .

\Rightarrow \frac{x}{1 - x} .

W {hat}^{'} s wrong in this ?

Commented by alex041103 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 by rahul 19 last updated on 10/Aug/18

ohh, yes u r right sir!