X x n xx

Any k k -th derivative for k < n k < n is going to have a limit of ∞ ∞ as x → ∞ x → ∞. Also, any derivative of ex e x is ex e x. Therefore, by applying L'Hospital successively, you will always have the form ∞∞ ∞ ∞, meaning that you can apply L'Hospital yet again. Eventually, you will have n! n! in the denominator and the ....

Hint: Use integration by parts. EDIT: Try several values of n n . ∫xexdx = (x − 1)ex + C ∫ x e x d x = ( x − 1) e x + C. ∫x2exdx = (x2 − 2x + 2)ex + C. ∫ x 2 e x d x = ( x 2 − 2 x + 2) e x + C. ∫x3exdx = (x3 − 3x2 + 6x − 6)ex + C. ∫ x 3 e x d x = ( x 3 − 3 x 2 + 6 x − 6) e x + C.Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. For math, science, nutrition, history ...Hint: Use integration by parts. EDIT: Try several values of n n . ∫xexdx = (x − 1)ex + C ∫ x e x d x = ( x − 1) e x + C. ∫x2exdx = (x2 − 2x + 2)ex + C. ∫ x 2 e x d x = ( x 2 − 2 x + 2) e x + C. ∫x3exdx = (x3 − 3x2 + 6x − 6)ex + C. ∫ x 3 e x d x = ( x 3 − 3 x 2 + 6 x − 6) e x + C.

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