sir Ma?h+t?que you have posted ∫(dx/(((x+1)....(x+n))^2 ))=......can you reposted it please


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Question Number 87692 by mind is power last updated on 05/Apr/20

$s i r M a ? h + t ? q u e y o u h a v e p o s t e d$ $\int \frac{d x}{{((x + 1) . . . . (x + n))}^{2}} = . . . . . . c a n y o u r e p o s t e d i t p l e a s e$
Commented by M±th+et£s last updated on 05/Apr/20

Commented by M±th+et£s last updated on 05/Apr/20

$y o u m e a e n t h i s s i r$
Commented by mind is power last updated on 05/Apr/20

$y e a h t h a n x$
Commented by M±th+et£s last updated on 05/Apr/20

$y o u a r e w e l c o m e s i r . h o p e y o u f i n d a s o l u t i o n$
Commented by mind is power last updated on 05/Apr/20

$i h a v e a n i d e a$ $\frac{\partial^{n}}{\partial a_{1} . . . . . \partial a_{n}} . \frac{1}{(x - a_{1}) . . . . . . . (x - a_{n})} = \frac{1}{\underset{k = 1}{\prod^{n}} {((x - a_{1}) . . . . . (x - a_{n}))}^{2}}$ $i w i l l p o s t s o l u t i o n a f t e r i f i n i s h i t s n o t s o e a s y$ $f o r r e d a c t i o n$
Commented by mind is power last updated on 06/Apr/20

$\frac{1}{(x - a_{0}) . . . . . (x - a_{n})} = \underset{k = 0}{\sum^{n}} \frac{1}{\underset{l = 0, l \neq k}{\prod^{n}} (a_{k} - a_{l}) (x - a_{k})} = f (a_{0}, . . a_{n})$ $\frac{\partial^{n + 1} f (a_{0}, . . . ., a_{n})}{\partial a_{0} . . . . \partial a_{n}} ∣_{(0, 1,, n)} = \frac{1}{(x^{2} . . . {(x - n)}^{2}} w e c a n s e e t h a t$ $\frac{\partial^{n + 1}}{\partial a_{0} . . . \partial_{a_{n}}} (\frac{1}{(x - a_{0}) . . . . . (x - a_{n})}) = \frac{1}{{(\underset{k = 0}{\prod^{n}} (x - a_{k}))}^{2}}$ $\partial^{n + 1} f (a_{0}, . . . . a_{n}) =$ $= - \underset{k = 0}{\sum^{n}} . \underset{j = 0, j \neq k}{\sum^{n}} \frac{2}{(a_{k} - a_{j}) . \underset{l = 1, l \neq k}{\prod^{n}} {(a_{k} - a_{l})}^{2} (x - a_{k})} + \underset{k = 0}{\sum^{n}} \frac{1}{\underset{l = 0, l \neq k}{\prod^{n}} {(a_{k} - a_{l})}^{2} {(x - a_{k})}^{2}}$ $\partial^{n + 1} f (0, 1, . . . . ., n) =$ $You can't use 'macro parameter character #' in math mode$ $You can't use 'macro parameter character #' in math mode$ $= \underset{k = 0}{\sum^{n}} \frac{2 {(n!)}^{2} (H_{n - k} - H_{k})}{{(k)}^{2} . . . (1) . . . . {(n - k)}^{2} . {(n!)}^{2}} \frac{1}{(x - k)} + \underset{k = 0}{\sum^{n}} \frac{{(n!)}^{2}}{{(n!)}^{2} {(k . . . 1 . . . (n - k))}^{2}} . \frac{1}{x - k}$ $= 2 \underset{k = 0}{\sum^{n}} \frac{{(\begin{matrix} n \\ k \end{matrix})}^{2} (H_{n - k} - H_{k})}{{(n!)}^{2}} . \frac{1}{x - k} + \frac{1}{{(n!)}^{2}} \underset{k = 0}{\sum^{n}} . \frac{{(\begin{matrix} n \\ k \end{matrix})}^{2}}{{(x - k)}^{2}}$ $\int \partial^{n + 1} f (0, . . . n) d x = \int \frac{d x}{x^{2} {(x - 1)}^{2} . . . {(x - n)}^{2}}$ $= 2 \underset{k = 0}{\sum^{n}} \frac{{(\begin{matrix} n \\ k \end{matrix})}^{2} (H_{n - k} - H_{k})}{{(n!)}^{2}} . \int \frac{1}{x - k} d x + \frac{1}{{(n!)}^{2}} \underset{k = 0}{\sum^{n}} \int . \frac{{(\begin{matrix} n \\ k \end{matrix})}^{2}}{{(x - k)}^{2}}$ $\frac{2}{{(n!)}^{2}} \underset{k = 0}{\sum^{n}} {(\begin{matrix} n \\ k \end{matrix})}^{2} (H_{n - k} - H_{k}) l n (x - k) + \frac{\underset{k = 0}{\sum^{n}} (\begin{matrix} n \\ k \end{matrix})}{{(n!)}^{2}} . \frac{1}{k - x}$ $Missing \left or extra \right$ $Σ$
Commented by M±th+et£s last updated on 06/Apr/20

$i a m s p e e c h l e s s s i r . g o d b l e s s y o u$
Commented by mind is power last updated on 06/Apr/20

$w i t h e p l e a s u r s i r, g o l d b l e s s Y o u t o o$ $i f y o u c s n r e p o s t e s o m m e u n s w e r d Q u a t i o n m a y$ $b e e i w i l l s e e s o m e s i d e a s$
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