1)calculate I_n = ∫_0 ^∞ (dx/((2x^2 +5x+3)^n )) 2) calculate ∫_0 ^∞ (dx/((2x^2 +5x+3)^2 )) and ∫


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Question Number 96834 by mathmax by abdo last updated on 05/Jun/20

$1) calculate I_{n} = \int_{0}^{\infty} \frac{dx}{{({2 x}^{2} + 5 x + 3)}^{n}}$ $2) calculate \int_{0}^{\infty} \frac{dx}{{({2 x}^{2} + 5 x + 3)}^{2}} and \int_{0}^{\infty} \frac{dx}{{({2 x}^{2} + 5 x + 3)}^{3}}$
Commented by mathmax by abdo last updated on 05/Jun/20

$n integr natural$
	Answered by abdomathmax last updated on 05/Jun/20
	$1) 2x^2 +5x+3 =0 →Δ =25−4.2.3 =1 ⇒ x_1 =((−5+1)/4) =−1 and x_2 =((−5−1)/4) =−(3/2) ⇒ 2x^2 +5x +3 =2(x+1)(x+(3/2)) =(x+1)(2x+3) ⇒ I_n =∫_0 ^∞ (dx/((x+1)^n (2x+3)^n )) =∫_0 ^∞ (dx/((((x+1)/(2x+3)))^n (2x+3)^(2n) )) changement ((x+1)/(2x+3)) =t give x+1 =2tx +3t ⇒ (1−2t)x =3t−1 ⇒x =((3t−1)/(1−2t)) ⇒ (dx/dt) = ((3(1−2t)−(3t−1)(−2))/((2t−1)^2 )) =((3−6t+6t−2)/((2t−1)^2 )) =(1/((2t−1)^2 )) also 2x+3 =((6t−2)/(1−2t)) +3 =((6t−2+3−6t)/(1−2t)) =(1/((1−2t))) ⇒ I_n =∫_(1/3) ^(1/2) (dt/((2t−1)^2 t^n ((1/(2t−1)))^(2n) )) =∫_(1/3) ^(1/2) (((2t−1)^(2n−2) )/t^n ) dt =∫_(1/3) ^(1/2) ((Σ_(k=0) ^(2n−2) C_(2n−2) ^k (2t)^k (−1)^(2n−2−k) )/t^n ) dt =Σ_(k=0) ^(2n−2) C_(2n−2) ^k 2^k (−1)^k ∫_(1/3) ^(1/2) t^(k−n) dt =Σ_(k=0 and k≠n−1) ^(2n−2) C_(2n−2) ^k (−2)^k [(1/(k−n+1))t^(k−n+1) ]_(1/3) ^(1/2) + C_(2n−2) ^(n−1) (−2)^(n−1) [ln∣t∣]_(1/3) ^(1/2) I_n =Σ_(k=0 and k≠n−1) ^(2n−2) ((C_(2n−2) ^k (−2)^k )/(k−n+1)){ (1/2^(k−n+1) )−(1/3^(k−n+1) )} +(−2)^(n−1) C_(2n−2) ^(n−1) {ln(3)−ln(2)}$
	$1) {2 x}^{2} + 5 x + 3 = 0 \to Δ = 25 - 4.2.3 = 1 \Rightarrow$ $x_{1} = \frac{- 5 + 1}{4} = - 1 and x_{2} = \frac{- 5 - 1}{4} = - \frac{3}{2} \Rightarrow$ ${2 x}^{2} + 5 x + 3 = 2 (x + 1) (x + \frac{3}{2}) = (x + 1) (2 x + 3) \Rightarrow$ $I_{n} = \int_{0}^{\infty} \frac{dx}{{(x + 1)}^{n} {(2 x + 3)}^{n}} = \int_{0}^{\infty} \frac{dx}{{(\frac{x + 1}{2 x + 3})}^{n} {(2 x + 3)}^{2 n}}$ $changement \frac{x + 1}{2 x + 3} = t give x + 1 = 2 tx + 3 t \Rightarrow$ $(1 - 2 t) x = 3 t - 1 \Rightarrow x = \frac{3 t - 1}{1 - 2 t} \Rightarrow$ $\frac{dx}{dt} = \frac{3 (1 - 2 t) - (3 t - 1) (- 2)}{{(2 t - 1)}^{2}} = \frac{3 - 6 t + 6 t - 2}{{(2 t - 1)}^{2}}$ $= \frac{1}{{(2 t - 1)}^{2}} also 2 x + 3 = \frac{6 t - 2}{1 - 2 t} + 3$ $= \frac{6 t - 2 + 3 - 6 t}{1 - 2 t} = \frac{1}{(1 - 2 t)} \Rightarrow$ $I_{n} = \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{dt}{{(2 t - 1)}^{2} t^{n} {(\frac{1}{2 t - 1})}^{2 n}}$ $= \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{{(2 t - 1)}^{2 n - 2}}{t^{n}} dt$ $= \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{\sum_{k = 0}^{2 n - 2} C_{2 n - 2}^{k} {(2 t)}^{k} {(- 1)}^{2 n - 2 - k}}{t^{n}} dt$ $= \sum_{k = 0}^{2 n - 2} C_{2 n - 2}^{k} 2^{k} {(- 1)}^{k} \int_{\frac{1}{3}}^{\frac{1}{2}} t^{k - n} dt$ $= \sum_{k = 0 and k \neq n - 1}^{2 n - 2} C_{2 n - 2}^{k} {(- 2)}^{k} {[\frac{1}{k - n + 1} t^{k - n + 1}]}_{\frac{1}{3}}^{\frac{1}{2}}$ $+ C_{2 n - 2}^{n - 1} {(- 2)}^{n - 1} {[\ln ∣ t ∣]}_{\frac{1}{3}}^{\frac{1}{2}}$ $I_{n} = \sum_{k = 0 and k \neq n - 1}^{2 n - 2} \frac{C_{2 n - 2}^{k} {(- 2)}^{k}}{k - n + 1} {\frac{1}{2^{k - n + 1}} - \frac{1}{3^{k - n + 1}}}$ $+ {(- 2)}^{n - 1} C_{2 n - 2}^{n - 1} {\ln (3) - \ln (2)}$
	Commented by abdomathmax last updated on 05/Jun/20
	$2) ∫_0 ^∞ (dx/((2x^2 +5x+3)^2 )) =I_2 (n=2) =Σ_(k=0 and k≠1) ^2 ((C_2 ^k (−2)^k )/(k−1)){(1/2^(k−1) )−(1/3^(k−1) )} −2 C_2 ^1 ln((3/2)) =−{ 2−3} +(4/1){(1/2)−(1/3)}−4ln((3/2)) =(5/3)−4ln((3/2))$
	$2) \int_{0}^{\infty} \frac{dx}{{({2 x}^{2} + 5 x + 3)}^{2}} = I_{2} (n = 2)$ $= \sum_{k = 0 and k \neq 1}^{2} \frac{C_{2}^{k} {(- 2)}^{k}}{k - 1} {\frac{1}{2^{k - 1}} - \frac{1}{3^{k - 1}}}$ $- 2 C_{2}^{1} \ln (\frac{3}{2})$ $= - {2 - 3} + \frac{4}{1} {\frac{1}{2} - \frac{1}{3}} - 4 \ln (\frac{3}{2})$ $= \frac{5}{3} - 4 \ln (\frac{3}{2})$
	Commented by abdomathmax last updated on 05/Jun/20
	$∫_0 ^∞ (dx/((2x^2 +5x +3)^3 )) =I_3 =Σ_(k=0 and k≠2) ^4 ((C_4 ^k (−2)^k )/(k−2)){(1/2^(k−2) )−(1/3^(k−2) )} +4 C_4 ^2 ln((3/2)) =−(1/2){4−9} +(((−2)C_4 ^1 )/(−1)){2−3} −((8C_4 ^3 )/1){(1/2)−(1/3)} +((16 C_4 ^4 )/2){(1/4)−(1/9)}+4C_4 ^2 ln((3/2)) =...$
	$\int_{0}^{\infty} \frac{dx}{{({2 x}^{2} + 5 x + 3)}^{3}} = I_{3}$ $= \sum_{k = 0 and k \neq 2}^{4} \frac{C_{4}^{k} {(- 2)}^{k}}{k - 2} {\frac{1}{2^{k - 2}} - \frac{1}{3^{k - 2}}}$ $+ 4 C_{4}^{2} \ln (\frac{3}{2})$ $= - \frac{1}{2} {4 - 9} + \frac{(- 2) C_{4}^{1}}{- 1} {2 - 3}$ $- \frac{{8 C}_{4}^{3}}{1} {\frac{1}{2} - \frac{1}{3}} + \frac{16 C_{4}^{4}}{2} {\frac{1}{4} - \frac{1}{9}} + {4 C}_{4}^{2} \ln (\frac{3}{2})$ $= . . .$
	Answered by Sourav mridha last updated on 05/Jun/20

	$1) I_{n} = \int_{0}^{\infty} \frac{d x}{{(2 x^{2} + 5 x + 3)}^{n}}$ $= \frac{1}{2^{n}} \int_{0}^{\infty} \frac{d x}{{(x + \frac{3}{2})}^{n} . {(x + 1)}^{n}}$ $n o w l e t [x + \frac{3}{2}] = k (x + 1)$ $a f t e r s o m e m a n i p u l a t i o n y o u g e t$ $t h i s n i c e r e s u l t . . .$ $= 2^{n - 1} \int_{1}^{\frac{3}{2}} \frac{{[k - 1]}^{2 (n - 1)}}{k^{n}} d k$ $n o w I_{2} = 2 \int_{1}^{\frac{3}{2}} \frac{{(k - 1)}^{2}}{k^{2}} d k = [\frac{5}{3} - 4 l n \frac{3}{2}]$ $a n d I_{3} = 2^{2} \int_{1}^{\frac{3}{2}} \frac{{(k - 1)}^{4}}{k^{3}} d k$ $= 4 \int_{1}^{\frac{3}{2}} \frac{\underset{r = 0}{\sum^{4}} {\overset{4}{C}}_{r} k^{(4 - r)} {(- 1)}^{r}}{k^{3}} dk$ $= - \frac{525}{54} + 24 l n \frac{3}{2}$ $★ v e r y l e n g t h y s o I a v o i d t h e$ $t o t a l c a l c u l a t i o n .$
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