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Question Number 45019 by rahul 19 last updated on 07/Oct/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 08/Oct/18
	$let t=1+(x/(1!))+(x^2 /(2∙!))+(x^3 /(3!))+...+(x^n /(n!)) (dt/dx)=0+1+(x/(1!))+(x^2 /(2!))+..+(x^(n−1) /((n−1)!))+(x^n /(n!))−(x^n /(n!)) (dt/dx)=t−(x^n /(n!)) (x^n /(n!))=t−(dt/dx) ((x^n dx)/(n!))=tdx−dt x^n dx=(tdx−dt)n! ∫(x^n /t)dx =n!∫((tdx−dt)/t) =n!∫dx−n!∫(dt/t) =(n!)x−n!lnt +c =n!(x−lnt)+c =n!{x−ln(1+x+(x^2 /(2!))+(x^3 /(3!))+...+(x^n /(n!)))}+c pls check$
	$l e t t = 1 + \frac{x}{1!} + \frac{x^{2}}{2 \cdot!} + \frac{x^{3}}{3!} + . . . + \frac{x^{n}}{n!}$ $\frac{d t}{d x} = 0 + 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + . . + \frac{x^{n - 1}}{(n - 1)!} + \frac{x^{n}}{n!} - \frac{x^{n}}{n!}$ $\frac{d t}{d x} = t - \frac{x^{n}}{n!}$ $\frac{x^{n}}{n!} = t - \frac{d t}{d x}$ $\frac{x^{n} d x}{n!} = t d x - d t$ $x^{n} d x = (t d x - d t) n!$ $\int \frac{x^{n}}{t} d x$ $= n! \int \frac{t d x - d t}{t}$ $= n! \int d x - n! \int \frac{d t}{t}$ $= (n!) x - n! l n t + c$ $= n! (x - l n t) + c$ $= n! {x - l n (1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . + \frac{x^{n}}{n!})} + c$ $p l s c h e c k$
	Commented by rahul 19 last updated on 08/Oct/18
	thank you so much sir ! ☺️�� Your ans. is correct i.e Option C.
	Commented by rahul 19 last updated on 08/Oct/18

	$H o w t h i s m e t h o d s t r i k e y o u r m i n d ?$ $I w a s s e a r c h i n g f o r s o m e s u b s t i t u t i o n$ $l i k e i n d e n o m i n a t o r i t h o u g h t i t i s$ $e x p a n s i o n o f e^{x} .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 08/Oct/18

	$i s t a r t e d t h i s p r o b l e m a t 1 - 30 a t n i g h t . . . a t f i r d t$ $i f o u n d b o t h f u n c t i o n i n t e r m s o f t a n d x . . .$ $b u t w h e n p l a c e d i n i n t r e g a l . . . i t g o t s e p a r a t e d . . .$
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