∫xe^x dx


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Question Number 75873 by Rio Michael last updated on 19/Dec/19

$\int {x e}^{x} d x$
	Answered by MJS last updated on 19/Dec/19

	$u^{'} = e^{x} \Rightarrow u = e^{x}$ $v = x \Rightarrow v^{'} = 1$ $\int u^{'} v = u v - \int {u v}^{'}$ $\int x e^{x} d x = x e^{x} - \int e^{x} d x = x e^{x} - e^{x} =$ $= (x - 1) e^{x} + C$
	Commented by Rio Michael last updated on 21/Dec/19

	$e x c u s e m e s i r a m k i n d a c o n f u s e d$ $w h e n i t c o m e s t o t h e s e b y p a r t s s t u f f s$ $w h y {c a n}^{'} t w e t a k e$ $u^{'} = x \Rightarrow u = \frac{1}{2} x^{2}$
	Commented by MJS last updated on 21/Dec/19

	$try it$ $u^{'} = x \Rightarrow u = \frac{1}{2} x^{2}$ $v = e^{x} \Rightarrow v^{'} = e^{x}$ $\int u^{'} v = u v - \int {u v}^{'}$ $\int x e^{x} d x = \frac{1}{2} x^{2} e^{x} - \frac{1}{2} \int x^{2} e^{x} d x$ $the point is, we must try to find u, v to make$ $it easier$
	Commented by Rio Michael last updated on 21/Dec/19

	$o k a y, s o w e j u s t c h o o s e t o m a k e o u r p r o b l e m e a s i e r .$
	Commented by MJS last updated on 21/Dec/19

	$yes . but it can be tricky sometimes$
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