calculate ∫_1 ^(√3) (dx/((x^2 +1)^2 (x+2)^5 ))


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Question Number 123710 by Bird last updated on 27/Nov/20

$c a l c u l a t e \int_{1}^{\sqrt{3}} \frac{d x}{{(x^{2} + 1)}^{2} {(x + 2)}^{5}}$
	Answered by MJS_new last updated on 27/Nov/20

	$Ostrogradski gives$ $\int \frac{d x}{{(x^{2} + 1)}^{2} {(x + 2)}^{5}} =$ $= - \frac{372 x^{5} + 2502 x^{4} + 6016 x^{3} + 6383 x^{2} + 4144 x + 3131}{7500 (x^{2} + 1) {(x + 2)}^{4}} -$ $- \frac{1}{625} \int \frac{31 x + 17}{(x^{2} + 1) (x + 2)} d x$ $- \frac{1}{625} \int \frac{31 x + 17}{(x^{2} + 1) (x + 2)} d x =$ $= - \frac{79}{6250} \int \frac{2 x}{x^{2} + 1} d x + \frac{3}{3125} \int \frac{d x}{x^{2} + 1} + \frac{79}{3125} \int \frac{d x}{x + 2} =$ $= - \frac{79}{6250} \ln (x^{2} + 1) + \frac{3}{3125} \arctan x + \frac{79}{3125} \ln ∣ x + 2 ∣ + C$ $\Rightarrow answer is$ $\frac{π}{12500} - \frac{146149}{81000} + \frac{2609 \sqrt{3}}{2500} + \frac{79}{6250} \ln \frac{7 + 4 \sqrt{3}}{18}$
	Commented by mathmax by abdo last updated on 28/Nov/20

	$thank you sir mjs$
	Answered by mathmax by abdo last updated on 28/Nov/20
	$complex method I =∫_1 ^(√3) (dx/((x−i)^2 (x+i)^2 (x+2)^5 )) =∫_1 ^(√3) (dx/((((x−i)/(x+i)))^2 (x+i)^4 (x+2)^5 )) we do the changement ((x−i)/(x+i))=t ⇒x−i=tx+it ⇒(1−t)x=it+i ⇒ x=i((1+t)/(1−t)) ⇒(dx/dt) =i((1−t−(1+t)(−1))/((1−t)^2 ))=i ((1−t+1+t)/((1−t)^2 ))=((2i)/((1−t)^2 )) x+i =((i+it)/(1−t)) +i =((i+it+i−it)/(1−t)) =((2i)/(1−t)) x+2 =((i+it)/(1−t)) +2 =((i+it+2−2t)/(1−t)) =(((−2+i)t +2+i)/(1−t)) ⇒ I =∫_((1−i)/(1+i)) ^(((√3)−i)/( (√3)+i)) ((2i)/((1−t)^2 t^2 (((2i)^4 )/((1−t)^4 ))((((−2+i)t+2+i)^5 )/((1−t)^5 ))))dt =((−1)/((2i)^3 ))∫_((1−i)/(1+i)) ^(((√3)−i)/( (√3)+i)) (((t−1)^9 )/((t−1)^2 t^2 {(−2+i)t +2+i}^5 ))dt =−(i/8)∫_((1−i)/(1+i)) ^(((√3)−i)/( (√3)+i)) (((t−1)^7 )/(t^2 {(−2+i)t+2+i}^5 ))dt =(i/8) ∫_((1−i)/(1+i)) ^(((√3)−i)/( (√3)+i)) ((Σ_(k=0) ^7 (−1)^k C_7 ^k t^k (−1)^(7−k) )/(t^2 {(−2+i)t+2+i}^5 ))dt rst decomposition....be continued...$
	$complex method$ $I = \int_{1}^{\sqrt{3}} \frac{dx}{{(x - i)}^{2} {(x + i)}^{2} {(x + 2)}^{5}} = \int_{1}^{\sqrt{3}} \frac{dx}{{(\frac{x - i}{x + i})}^{2} {(x + i)}^{4} {(x + 2)}^{5}}$ $we do the changement \frac{x - i}{x + i} = t \Rightarrow x - i = tx + it \Rightarrow (1 - t) x = it + i \Rightarrow$ $x = i \frac{1 + t}{1 - t} \Rightarrow \frac{dx}{dt} = i \frac{1 - t - (1 + t) (- 1)}{{(1 - t)}^{2}} = i \frac{1 - t + 1 + t}{{(1 - t)}^{2}} = \frac{2 i}{{(1 - t)}^{2}}$ $x + i = \frac{i + it}{1 - t} + i = \frac{i + it + i - it}{1 - t} = \frac{2 i}{1 - t}$ $x + 2 = \frac{i + it}{1 - t} + 2 = \frac{i + it + 2 - 2 t}{1 - t} = \frac{(- 2 + i) t + 2 + i}{1 - t} \Rightarrow$ $I = \int_{\frac{1 - i}{1 + i}}^{\frac{\sqrt{3} - i}{\sqrt{3} + i}} \frac{2 i}{{(1 - t)}^{2} t^{2} \frac{{(2 i)}^{4}}{{(1 - t)}^{4}} \frac{{((- 2 + i) t + 2 + i)}^{5}}{{(1 - t)}^{5}}} dt$ $= \frac{- 1}{{(2 i)}^{3}} \int_{\frac{1 - i}{1 + i}}^{\frac{\sqrt{3} - i}{\sqrt{3} + i}} \frac{{(t - 1)}^{9}}{{(t - 1)}^{2} t^{2} {(- 2 + i) t + 2 + i}^{5}} dt$ $= - \frac{i}{8} \int_{\frac{1 - i}{1 + i}}^{\frac{\sqrt{3} - i}{\sqrt{3} + i}} \frac{{(t - 1)}^{7}}{t^{2} {(- 2 + i) t + 2 + i}^{5}} dt$ $= \frac{i}{8} \int_{\frac{1 - i}{1 + i}}^{\frac{\sqrt{3} - i}{\sqrt{3} + i}} \frac{\sum_{k = 0}^{7} {(- 1)}^{k} C_{7}^{k} t^{k} {(- 1)}^{7 - k}}{t^{2} {(- 2 + i) t + 2 + i}^{5}} dt$ $rst decomposition . . . . be continued . . .$
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