calculate ∫_1 ^∞ (dx/((x+1)^3 (x^2 +1)^2 ))


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Question Number 88422 by abdomathmax last updated on 10/Apr/20

$c a l c u l a t e \int_{1}^{\infty} \frac{d x}{{(x + 1)}^{3} {(x^{2} + 1)}^{2}}$
Commented by mathmax by abdo last updated on 12/Apr/20
$first let find I = ∫ (dx/((x+1)^3 (x^2 +1)^2 )) ⇒ I =∫ (dx/((x+1)^3 (x−i)^2 (x+i)^2 )) =∫ (dx/((x+1)^3 (((x−i)/(x+i)))^2 (x+i)^4 )) changement ((x−i)/(x+i)) =t give x−i =tx+it ⇒(1−t)x =i(1+t) x =i((1+t)/(1−t)) ⇒(dx/dt) =i×((1−t+(1+t))/((1−t)^2 )) =((2i)/((t−1)^2 )) x+1 =((i+it)/(1−t))+1 =((i+it+1−t)/(1−t)) =(((−1+i)t+i+1)/(1−t)) x+i =((i+it)/(1−t))+i =((i+it+i−it)/(1−t)) =((2i)/(1−t)) ⇒= I =∫ ((2idt)/((t−1)^2 ((((i−1)t +i+1)/(1−t)))^3 t^2 (((2i)/(1−t)))^4 )) =−(1/((2i)^3 ))∫ (((t−1)^7 dt)/((t−1)^2 {(i−1)t +i+1}^3 )) =(1/(8i)) ∫ (((t−1)^5 )/({(i−1)t +i+1)^3 ))dt =(1/(8i(i−1)^3 ))∫ (((t−1)^5 )/((t +((i+1)/(i−1)))^3 ))dt =(1/(8i(i−1)^3 ))∫ (((t−1)^5 )/((t−i)^3 ))dt ⇒ 8i(i−1)^3 I =∫ ((Σ_(k=0) ^5 C_5 ^k t^k (−1)^(5−k) )/((t−i)^3 ))dt =−Σ_(k=0) ^5 (−1)^k C_5 ^k ∫ (t^k /((t−i)^3 ))dt =−C_5 ^0 ∫ (dt/((t−i)^3 )) +C_5 ^1 ∫ (t/((t−i)^3 ))dt −C_5 ^2 ∫ (t^2 /((t−i)^3 ))dt+... +C_5 ^5 ∫ (t^5 /((t−i)^3 ))dt .....be continued....$
$f i r s t l e t f i n d I = \int \frac{d x}{{(x + 1)}^{3} {(x^{2} + 1)}^{2}} \Rightarrow$ $I = \int \frac{d x}{{(x + 1)}^{3} {(x - i)}^{2} {(x + i)}^{2}} = \int \frac{d x}{{(x + 1)}^{3} {(\frac{x - i}{x + i})}^{2} {(x + i)}^{4}}$ $c h a n g e m e n t \frac{x - i}{x + i} = t g i v e x - i = t x + i t \Rightarrow (1 - t) x = i (1 + t)$ $x = i \frac{1 + t}{1 - t} \Rightarrow \frac{d x}{d t} = i \times \frac{1 - t + (1 + t)}{{(1 - t)}^{2}} = \frac{2 i}{{(t - 1)}^{2}}$ $x + 1 = \frac{i + i t}{1 - t} + 1 = \frac{i + i t + 1 - t}{1 - t} = \frac{(- 1 + i) t + i + 1}{1 - t}$ $x + i = \frac{i + i t}{1 - t} + i = \frac{i + i t + i - i t}{1 - t} = \frac{2 i}{1 - t} \Rightarrow=$ $I = \int \frac{2 i d t}{{(t - 1)}^{2} {(\frac{(i - 1) t + i + 1}{1 - t})}^{3} t^{2} {(\frac{2 i}{1 - t})}^{4}}$ $= - \frac{1}{{(2 i)}^{3}} \int \frac{{(t - 1)}^{7} d t}{{(t - 1)}^{2} {(i - 1) t + i + 1}^{3}}$ $= \frac{1}{8 i} \int \frac{{(t - 1)}^{5}}{{{(i - 1) t + i + 1)}^{3}} d t = \frac{1}{8 i {(i - 1)}^{3}} \int \frac{{(t - 1)}^{5}}{{(t + \frac{i + 1}{i - 1})}^{3}} d t$ $= \frac{1}{8 i {(i - 1)}^{3}} \int \frac{{(t - 1)}^{5}}{{(t - i)}^{3}} d t \Rightarrow$ $8 i {(i - 1)}^{3} I = \int \frac{\sum_{k = 0}^{5} C_{5}^{k} t^{k} {(- 1)}^{5 - k}}{{(t - i)}^{3}} d t$ $= - \sum_{k = 0}^{5} {(- 1)}^{k} C_{5}^{k} \int \frac{t^{k}}{{(t - i)}^{3}} d t$ $= - C_{5}^{0} \int \frac{d t}{{(t - i)}^{3}} + C_{5}^{1} \int \frac{t}{{(t - i)}^{3}} d t - C_{5}^{2} \int \frac{t^{2}}{{(t - i)}^{3}} d t + . . .$ $+ C_{5}^{5} \int \frac{t^{5}}{{(t - i)}^{3}} d t . . . . . b e c o n t i n u e d . . . .$
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