find ∫ (dx/((x+1)^3 (x^2 +3)^2 ))


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Question Number 81432 by abdomathmax last updated on 13/Feb/20

$f i n d \int \frac{d x}{{(x + 1)}^{3} {(x^{2} + 3)}^{2}}$
Commented by abdomathmax last updated on 20/Feb/20
$let try complex method I =∫ (dx/((x+1)^3 (x^2 +3)^2 ))dx ⇒I=∫ (dx/((x+1)^3 (x−i(√3))^2 (x+i(√3))^2 )) =∫ (dx/((((x+1)/(x+i(√3))))^3 ×(x+i(√3))^5 (x−i(√3))^2 )) changement ((x+1)/(x+i(√3))) =t give x+1=tx+i(√3)t ⇒ (1−t)x=i(√3)t−1 ⇒x=((i(√3)t−1)/(1−t)) dx =((i(√3)(1−t)−(i(√3)t−1)(−1))/((1−t)^2 ))dt =((i(√3)−i(√3)t+i(√3)t−1)/((t−1)^2 ))dt =((i(√3)−1)/((t−1)^2 ))dt x+i(√3)=((i(√3)t−1)/(1−t)) +i(√3)=((i(√3)t−1+i(√3)−i(√3)t)/(1−t)) =((i(√3)−1)/(1−t)) =((1−i(√3))/(t−1)) x−i(√3)=((i(√3)t−1)/(1−t))−i(√3)=((i(√3)t−1−i(√3)+i(√3)t)/((1−t))) =((2i(√3)t−1−i(√3))/((1−t))) =((−2i(√3)t+1+i(√3))/((t−1))) ⇒ I =(i(√3)−1) ∫ (dt/((t−1)^2 t^3 (((1−i(√3))/(t−1)))^5 (((−2i(√3)t+1+i(√3))/(t−1)))^2 )) =(i(√3)−1) ∫ (((t−1)^7 dt)/((t−1)^2 t^3 (−2i(√3)t +1+i(√3))^2 (1−i(√3))^5 )) =−(1/((i(√3)−1)^4 )) ∫ (((t−1)^5 )/(t^3 (−2i(√3)t +i(√3))^2 ))dt =(1/(3(i(√3)−1)^4 )) ∫ (((t−1)^5 )/(t^2 (−2t+1)^2 ))dt =λ ∫ ((Σ_(k=0) ^5 C_5 ^k t^k (−1)^(5−k) )/(t^2 (2t−1)^2 ))dt (λ=(1/(3(i(√3)−1)^4 ))) =λ∫ ((− C_5 ^0 +C_5 ^1 t−C_5 ^2 t^2 +C_5 ^3 t^3 −C_5 ^4 t^4 +C_5 ^5 t^5 )/(t^2 (2t−1)^2 ))dt =λ ∫ {−(C_5 ^0 /(t^2 (2t−1)^2 ))+(C_5 ^1 /(t(2t−1)^2 ))−(C_5 ^2 /((2t−1)^2 )) +((C_5 ^3 t)/((2t−1)^2 )) −((C_5 ^4 t^2 )/((2t−1)^2 )) +((C_5 ^5 t^5 )/((2t−1)^2 ))}dt ...be continued...$
$l e t t r y c o m p l e x m e t h o d I = \int \frac{d x}{{(x + 1)}^{3} {(x^{2} + 3)}^{2}} d x \Rightarrow I = \int \frac{d x}{{(x + 1)}^{3} {(x - i \sqrt{3})}^{2} {(x + i \sqrt{3})}^{2}}$ $= \int \frac{d x}{{(\frac{x + 1}{x + i \sqrt{3}})}^{3} \times {(x + i \sqrt{3})}^{5} {(x - i \sqrt{3})}^{2}}$ $c h a n g e m e n t \frac{x + 1}{x + i \sqrt{3}} = t g i v e x + 1 = t x + i \sqrt{3} t \Rightarrow$ $(1 - t) x = i \sqrt{3} t - 1 \Rightarrow x = \frac{i \sqrt{3} t - 1}{1 - t}$ $d x = \frac{i \sqrt{3} (1 - t) - (i \sqrt{3} t - 1) (- 1)}{{(1 - t)}^{2}} d t$ $= \frac{i \sqrt{3} - i \sqrt{3} t + i \sqrt{3} t - 1}{{(t - 1)}^{2}} d t = \frac{i \sqrt{3} - 1}{{(t - 1)}^{2}} d t$ $x + i \sqrt{3} = \frac{i \sqrt{3} t - 1}{1 - t} + i \sqrt{3} = \frac{i \sqrt{3} t - 1 + i \sqrt{3} - i \sqrt{3} t}{1 - t}$ $= \frac{i \sqrt{3} - 1}{1 - t} = \frac{1 - i \sqrt{3}}{t - 1}$ $x - i \sqrt{3} = \frac{i \sqrt{3} t - 1}{1 - t} - i \sqrt{3} = \frac{i \sqrt{3} t - 1 - i \sqrt{3} + i \sqrt{3} t}{(1 - t)}$ $= \frac{2 i \sqrt{3} t - 1 - i \sqrt{3}}{(1 - t)} = \frac{- 2 i \sqrt{3} t + 1 + i \sqrt{3}}{(t - 1)} \Rightarrow$ $I = (i \sqrt{3} - 1) \int \frac{d t}{{(t - 1)}^{2} t^{3} {(\frac{1 - i \sqrt{3}}{t - 1})}^{5} {(\frac{- 2 i \sqrt{3} t + 1 + i \sqrt{3}}{t - 1})}^{2}}$ $= (i \sqrt{3} - 1) \int \frac{{(t - 1)}^{7} d t}{{(t - 1)}^{2} t^{3} {(- 2 i \sqrt{3} t + 1 + i \sqrt{3})}^{2} {(1 - i \sqrt{3})}^{5}}$ $= - \frac{1}{{(i \sqrt{3} - 1)}^{4}} \int \frac{{(t - 1)}^{5}}{t^{3} {(- 2 i \sqrt{3} t + i \sqrt{3})}^{2}} d t$ $= \frac{1}{3 {(i \sqrt{3} - 1)}^{4}} \int \frac{{(t - 1)}^{5}}{t^{2} {(- 2 t + 1)}^{2}} d t$ $= λ \int \frac{\sum_{k = 0}^{5} C_{5}^{k} t^{k} {(- 1)}^{5 - k}}{t^{2} {(2 t - 1)}^{2}} d t (λ = \frac{1}{3 {(i \sqrt{3} - 1)}^{4}})$ $= λ \int \frac{- C_{5}^{0} + C_{5}^{1} t - C_{5}^{2} t^{2} + C_{5}^{3} t^{3} - C_{5}^{4} t^{4} + C_{5}^{5} t^{5}}{t^{2} {(2 t - 1)}^{2}} d t$ $= λ \int {- \frac{C_{5}^{0}}{t^{2} {(2 t - 1)}^{2}} + \frac{C_{5}^{1}}{t {(2 t - 1)}^{2}} - \frac{C_{5}^{2}}{{(2 t - 1)}^{2}} + \frac{C_{5}^{3} t}{{(2 t - 1)}^{2}}$ $- \frac{C_{5}^{4} t^{2}}{{(2 t - 1)}^{2}} + \frac{C_{5}^{5} t^{5}}{{(2 t - 1)}^{2}}} d t . . . b e c o n t i n u e d . . .$
	Answered by MJS last updated on 13/Feb/20

	$\int \frac{d x}{{(x + 1)}^{3} {(x^{2} + 3)}^{2}} =$ $Q_{1} (x) = {(x + 1)}^{2} (x^{2} + 3)$ $Q_{2} (x) = (x + 1) (x^{2} + 3)$ $P_{1} (x) = - \frac{1}{12} x^{3} - \frac{13}{96} x^{2} - \frac{5}{24} x - \frac{9}{32}$ $P_{2} (x) = - \frac{x}{12} - \frac{1}{48}$ $= - \frac{8 x^{3} + 13 x^{2} + 20 x + 27}{96 {(x + 1)}^{2} (x^{2} + 3)} - \frac{1}{48} \int \frac{4 x + 1}{(x + 1) (x^{2} + 3)} d x$ $\int \frac{4 x + 1}{(x + 1) (x^{2} + 3)} d x = - \frac{3}{4} \int \frac{d x}{x + 1} + \frac{1}{4} \int \frac{3 x + 13}{x^{2} + 3} d x =$ $= - \frac{3}{4} \ln ∣ x + 1 ∣ + \frac{3}{8} \ln (x^{2} + 3) + \frac{13 \sqrt{3}}{12} \arctan \frac{\sqrt{3} x}{3} =$ $= \frac{3}{8} \ln \frac{x^{2} + 3}{{(x + 1)}^{2}} + \frac{13 \sqrt{3}}{12} \arctan \frac{\sqrt{3} x}{3}$ $= - \frac{8 x^{3} + 13 x^{2} + 20 x + 27}{96 {(x + 1)}^{2} (x^{2} + 3)} - \frac{1}{128} \ln \frac{x^{2} + 3}{{(x + 1)}^{2}} - \frac{13 \sqrt{3}}{576} \arctan \frac{\sqrt{3} x}{3} + C$
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