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Question Number 76317 by Master last updated on 26/Dec/19

Commented by john santu last updated on 26/Dec/19

$\int e^{x l n x} d x . w i t h i n t e g r a t i o n b y p a r t$ $u = e^{x l n x} \to d u = (l n x + 1) e^{x l n x} . d v = d x$ $I = {x e}^{x l n x} - \int (x l n x + x) e^{x l n x} d x$
Commented by turbo msup by abdo last updated on 26/Dec/19

$a t f o r m o f s e r i e$ $\int x^{x} d x = \int e^{x l n (x)} d x$ $= \int \sum_{n = 0}^{\infty} \frac{x^{n} {(l n x)}^{n}}{n!} d x$ $= \sum_{n = 0}^{\infty} \frac{1}{n!} \int x^{n} {(l n x)}^{n} d x$ $a n d a_{n} = \int x^{n} {(l n x)}^{n} d x c a n b e$ $f i u n d b y r e c u r r e n c e . . .$
Commented by mr W last updated on 27/Dec/19

$i s e e a b s o l u t e l y n o s e n s e f o r$ $r e p e a t e d p o s t i n g q u e s t i o n s l i k e$ $\int x^{x} d x o r \int x! d x .$
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