1)find ∫ (dx/((x^2 −1)^9 )) 2) calculate ∫


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Question Number 82970 by mathmax by abdo last updated on 26/Feb/20

$1) f i n d \int \frac{d x}{{(x^{2} - 1)}^{9}}$ $2) c a l c u l a t e \int_{2}^{+ \infty} \frac{d x}{{(x^{2} - 1)}^{9}}$
Commented by mathmax by abdo last updated on 26/Feb/20
$1)let I =∫ (dx/((x^2 −1)^9 )) ⇒I =∫ (dx/((x−1)^9 (x+1)^9 )) =∫ (dx/((((x−1)/(x+1)))^9 (x+1)^(18) )) changement ((x−1)/(x+1)) =t give x−1=tx+t ⇒(1−t)x=t+1 ⇒x=((t+1)/(1−t)) dx =((1−t+(t+1))/((1−t)^2 ))dt =(2/((1−t)^2 ))dt and x+1=((t+1)/(1−t))+1 =((t+1+1−t)/(1−t))=(2/(1−t)) I =∫ (2/((t−1)^2 .t^9 .((2/(1−t)))^(18) ))dt =(1/2^(17) )∫ (((t−1)^(18) )/((t−1)^2 t^9 ))dt =(1/2^(17) )∫ (((t−1)^(16) )/t^9 )dt =(1/2^(17) )∫ ((Σ_(k=0) ^(16) C_(16) ^k t^k (−1)^(16−k) )/t^9 )dt 2^(17) I =Σ_(k=0) ^(16) (−1)^k C_(16) ^k ∫ t^(k−9) dt =Σ_(k=0 and k≠8) ^(16) (−1)^k C_(16) ^k ×(1/(k−8))t^(k−8) +C_(16) ^8 ln∣t∣ +C =Σ_(k=0 and k≠8) ^(16) (((−1)^k )/(k−8))C_(16) ^k (((x−1)/(x+1)))^(k−8) +C_(16) ^8 ln∣((x−1)/(x+1))∣ +C ⇒ I =(1/2^(17) ){ Σ_(k=0and k≠8) ^(16) (((−1)^k )/(k−8))C_(16) ^k (((x−1)/(x+1)))^(k−8) +C_(16) ^8 ln∣((x−1)/(x+1))∣} +C$
$1) l e t I = \int \frac{d x}{{(x^{2} - 1)}^{9}} \Rightarrow I = \int \frac{d x}{{(x - 1)}^{9} {(x + 1)}^{9}} = \int \frac{d x}{{(\frac{x - 1}{x + 1})}^{9} {(x + 1)}^{18}}$ $c h a n g e m e n t \frac{x - 1}{x + 1} = t g i v e x - 1 = t x + t \Rightarrow (1 - t) x = t + 1 \Rightarrow x = \frac{t + 1}{1 - t}$ $d x = \frac{1 - t + (t + 1)}{{(1 - t)}^{2}} d t = \frac{2}{{(1 - t)}^{2}} d t a n d x + 1 = \frac{t + 1}{1 - t} + 1 = \frac{t + 1 + 1 - t}{1 - t} = \frac{2}{1 - t}$ $I = \int \frac{2}{{(t - 1)}^{2} . t^{9} . {(\frac{2}{1 - t})}^{18}} d t = \frac{1}{2^{17}} \int \frac{{(t - 1)}^{18}}{{(t - 1)}^{2} t^{9}} d t$ $= \frac{1}{2^{17}} \int \frac{{(t - 1)}^{16}}{t^{9}} d t = \frac{1}{2^{17}} \int \frac{\sum_{k = 0}^{16} C_{16}^{k} t^{k} {(- 1)}^{16 - k}}{t^{9}} d t$ $2^{17} I = \sum_{k = 0}^{16} {(- 1)}^{k} C_{16}^{k} \int t^{k - 9} d t$ $= \sum_{k = 0 a n d k \neq 8}^{16} {(- 1)}^{k} C_{16}^{k} \times \frac{1}{k - 8} t^{k - 8} + C_{16}^{8} l n ∣ t ∣ + C$ $= \sum_{k = 0 a n d k \neq 8}^{16} \frac{{(- 1)}^{k}}{k - 8} C_{16}^{k} {(\frac{x - 1}{x + 1})}^{k - 8} + C_{16}^{8} l n ∣ \frac{x - 1}{x + 1} ∣ + C \Rightarrow$ $I = \frac{1}{2^{17}} {\sum_{k = 0 a n d k \neq 8}^{16} \frac{{(- 1)}^{k}}{k - 8} C_{16}^{k} {(\frac{x - 1}{x + 1})}^{k - 8} + C_{16}^{8} l n ∣ \frac{x - 1}{x + 1} ∣} + C$
Commented by mathmax by abdo last updated on 26/Feb/20
$2) ∫_2 ^(+∞) (dx/((x^2 −1)^9 ))=(1/2^(17) )[Σ_(k=0 and k≠8) ^(16) (((−1)^k )/(k−8))C_(16) ^k (((x−1)/(x+1)))^(k−8) +C_(16) ^8 ln∣((x−1)/(x+1))∣]_2 ^(+∞) =(1/2^(17) ){Σ_(k=0 and k≠8) ^(16) (((−1)^k )/(k−8))C_(16) ^k −Σ_(k=0 and k≠8) ^(16) (((−1)^k )/(k−8))C_(16) ^k ((1/3))^(k−8) −C_(16) ^8 ln((1/3))} =(1/2^(17) )Σ_(k=0 and k≠8) ^(16) (((−1)^k )/(k−8))C_(16) ^k (1−(1/3^(k−8) ))+(C_(16) ^8 /2^(17) )ln(3)$
$2) \int_{2}^{+ \infty} \frac{d x}{{(x^{2} - 1)}^{9}} = \frac{1}{2^{17}} {[\sum_{k = 0 a n d k \neq 8}^{16} \frac{{(- 1)}^{k}}{k - 8} C_{16}^{k} {(\frac{x - 1}{x + 1})}^{k - 8} + C_{16}^{8} l n ∣ \frac{x - 1}{x + 1} ∣]}_{2}^{+ \infty}$ $= \frac{1}{2^{17}} {\sum_{k = 0 a n d k \neq 8}^{16} \frac{{(- 1)}^{k}}{k - 8} C_{16}^{k} - \sum_{k = 0 a n d k \neq 8}^{16} \frac{{(- 1)}^{k}}{k - 8} C_{16}^{k} {(\frac{1}{3})}^{k - 8}$ $- C_{16}^{8} l n (\frac{1}{3})}$ $= \frac{1}{2^{17}} \sum_{k = 0 a n d k \neq 8}^{16} \frac{{(- 1)}^{k}}{k - 8} C_{16}^{k} (1 - \frac{1}{3^{k - 8}}) + \frac{C_{16}^{8}}{2^{17}} l n (3)$
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