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Question Number 159172 by aliibrahim1 last updated on 13/Nov/21

	Answered by ajfour last updated on 13/Nov/21
	$Apply Integration by parts;also ∫e^x {f(x)+f ′(x)}dx=e^x f(x)+c Here (y/((y+1)^2 ))=((y+1−1)/((y+1)^2 )) = (1/(y+1))+((−1)/(1+y^2 ))=f(y)+f ′(y) hence ∫((e^y ydy)/((y+1)^2 ))=(e^y /(y+1))+c$
	$A p p l y I n t e g r a t i o n b y p a r t s; a l s o$ $\int e^{x} {f (x) + f^{'} (x)} d x = e^{x} f (x) + c$ $H e r e \frac{y}{{(y + 1)}^{2}} = \frac{y + 1 - 1}{{(y + 1)}^{2}}$ $= \frac{1}{y + 1} + \frac{- 1}{1 + y^{2}} = f (y) + f^{'} (y)$ $h e n c e \int \frac{e^{y} y d y}{{(y + 1)}^{2}} = \frac{e^{y}}{y + 1} + c$
	Commented by aliibrahim1 last updated on 13/Nov/21

	$t h x s i r$
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