calculate A_λ =∫_3 ^∞ (dx/((x+λ)^3 (x−2)^4 )) (λ>0)


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Question Number 85641 by abdomathmax last updated on 23/Mar/20

$c a l c u l a t e A_{λ} = \int_{3}^{\infty} \frac{d x}{{(x + λ)}^{3} {(x - 2)}^{4}} (λ > 0)$
Commented bymathmax by abdo last updated on 24/Mar/20
$A_λ =∫_3 ^(+∞) (dx/((((x+λ)/(x−2)))^3 (x−2)^7 )) changement ((x+λ)/(x−2))=t give x+λ =tx−2t ⇒(1−t)x=−2t−λ ⇒x=((−2t−λ)/(1−t)) =((2t+λ)/(t−1)) ⇒ (dx/dt) =((2(t−1)−(2t+λ))/((t−1)^2 )) =((2t−2−2t−λ)/((t−1)^2 )) =((−2−λ)/((t−1)^2 )) x−2 =((2t+λ)/(t−1))−2 =((2t+λ−2t+2)/(t−1)) =((λ+2)/(t−1)) ⇒ A_λ =− ∫_(3+λ) ^1 (1/(t^3 (((λ+2)/(t−1)))^7 ))×((λ+2)/((t−1)^2 ))dt =(1/((λ+2)^6 ))∫_1 ^(3+λ) (dt/(t^3 (((λ+2)/(t−1)))^(−5) )) =(1/((λ+2)))∫_1 ^(3+λ) (((t−1)^5 )/t^3 )dt ⇒ (λ+2)A_λ =∫_1 ^(3+λ) ((Σ_(k=0) ^5 C_5 ^k t^k (−1)^(5−k) )/t^3 )dt =−∫_1 ^(3+λ) Σ_(k=0) ^5 (−1)^k C_5 ^k t^(k−3) dt =−Σ_(k=0) ^5 (−1)^k C_5 ^k ∫_1 ^(3+λ) t^(k−3) dt =−Σ_(k=0 and k≠2) ^5 (−1)^k C_5 ^k [(1/(k−2))t^(k−2) ]_1 ^(3+λ) −C_5 ^2 ∫_1 ^(3+λ) (dt/t) =−Σ_(k=0) ^5 (((−1)^k C_5 ^k )/(k−2)){ (3+λ)^(k−2) −1}−C_5 ^2 {ln∣3+λ∣} ⇒ A_λ =−(1/(λ+2))Σ_(k=0and k≠2) ^5 (((−1)^k C_5 ^k )/(k−2)){(3+λ)^(k−2) −1} −(1/(λ+2)) C_5 ^2 ln∣3+λ∣ .$
$A_{λ} = \int_{3}^{+ \infty} \frac{d x}{{(\frac{x + λ}{x - 2})}^{3} {(x - 2)}^{7}} c h a n g e m e n t \frac{x + λ}{x - 2} = t g i v e$ $x + λ = t x - 2 t \Rightarrow (1 - t) x = - 2 t - λ \Rightarrow x = \frac{- 2 t - λ}{1 - t} = \frac{2 t + λ}{t - 1} \Rightarrow$ $\frac{d x}{d t} = \frac{2 (t - 1) - (2 t + λ)}{{(t - 1)}^{2}} = \frac{2 t - 2 - 2 t - λ}{{(t - 1)}^{2}} = \frac{- 2 - λ}{{(t - 1)}^{2}}$ $x - 2 = \frac{2 t + λ}{t - 1} - 2 = \frac{2 t + λ - 2 t + 2}{t - 1} = \frac{λ + 2}{t - 1} \Rightarrow$ $A_{λ} = - \int_{3 + λ}^{1} \frac{1}{t^{3} {(\frac{λ + 2}{t - 1})}^{7}} \times \frac{λ + 2}{{(t - 1)}^{2}} d t$ $= \frac{1}{{(λ + 2)}^{6}} \int_{1}^{3 + λ} \frac{d t}{t^{3} {(\frac{λ + 2}{t - 1})}^{- 5}} = \frac{1}{(λ + 2)} \int_{1}^{3 + λ} \frac{{(t - 1)}^{5}}{t^{3}} d t \Rightarrow$ $(λ + 2) A_{λ} = \int_{1}^{3 + λ} \frac{\sum_{k = 0}^{5} C_{5}^{k} t^{k} {(- 1)}^{5 - k}}{t^{3}} d t$ $= - \int_{1}^{3 + λ} \sum_{k = 0}^{5} {(- 1)}^{k} C_{5}^{k} t^{k - 3} d t$ $= - \sum_{k = 0}^{5} {(- 1)}^{k} C_{5}^{k} \int_{1}^{3 + λ} t^{k - 3} d t$ $= - \sum_{k = 0 a n d k \neq 2}^{5} {(- 1)}^{k} C_{5}^{k} {[\frac{1}{k - 2} t^{k - 2}]}_{1}^{3 + λ} - C_{5}^{2} \int_{1}^{3 + λ} \frac{d t}{t}$ $= - \sum_{k = 0}^{5} \frac{{(- 1)}^{k} C_{5}^{k}}{k - 2} {{(3 + λ)}^{k - 2} - 1} - C_{5}^{2} {l n ∣ 3 + λ ∣} \Rightarrow$ $A_{λ} = - \frac{1}{λ + 2} \sum_{k = 0 a n d k \neq 2}^{5} \frac{{(- 1)}^{k} C_{5}^{k}}{k - 2} {{(3 + λ)}^{k - 2} - 1}$ $- \frac{1}{λ + 2} C_{5}^{2} l n ∣ 3 + λ ∣ .$
Commented bymathmax by abdo last updated on 24/Mar/20

$λ \neq - 2$
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