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Question Number 104197 by mathmax by abdo last updated on 20/Jul/20

calculate ∫_5 ^(+∞) (dx/((x^2 −9)^4 ))

$$\mathrm{calculate}\:\int_{\mathrm{5}} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{4}} } \\ $$

Answered by Dwaipayan Shikari last updated on 20/Jul/20

∫_5 ^(+∞) ((2xdx)/(2x(x^2 −9)^4 ))=(1/2)∫_(16) ^(+∞) (dt/(√(9+t))).(1/t^4 ) {t=9tan^2 θ, 1=18tanθsec^2 θ(dθ/dt) (1/6)∫_(tan^(−1) (4/3)) ^(π/2) ((18tanθsec^2 θdθ)/(secθ.9^4 tan^8 θ)) {take x^2 −9=t (1/3).(1/9^3 )∫_(tan^(−1) (4/3)) ^(π/2) ((secθ)/(tan^7 θ))dθ.....continue

$$\int_{\mathrm{5}} ^{+\infty} \frac{\mathrm{2xdx}}{\mathrm{2x}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{16}} ^{+\infty} \frac{\mathrm{dt}}{\sqrt{\mathrm{9}+\mathrm{t}}}.\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{4}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{\mathrm{t}=\mathrm{9tan}^{\mathrm{2}} \theta,\:\mathrm{1}=\mathrm{18tan}\theta\mathrm{sec}^{\mathrm{2}} \theta\frac{\mathrm{d}\theta}{\mathrm{dt}}\right. \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{18tan}\theta\mathrm{sec}^{\mathrm{2}} \theta\mathrm{d}\theta}{\mathrm{sec}\theta.\mathrm{9}^{\mathrm{4}} \mathrm{tan}^{\mathrm{8}} \theta}\:\:\:\left\{\mathrm{take}\:\mathrm{x}^{\mathrm{2}} −\mathrm{9}=\mathrm{t}\right. \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}.\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{3}} }\int_{\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sec}\theta}{\mathrm{tan}^{\mathrm{7}} \theta}\mathrm{d}\theta.....\mathrm{continue} \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 20/Jul/20

A =∫_5 ^∞ (dx/((x^2 −9)^4 )) ⇒A =∫_5 ^∞ (dx/((x−3)^4 (x+3)^4 )) =∫_5 ^∞ (dx/((((x−3)/(x+3)))^4 (x+3)^8 )) we do the changement ((x−3)/(x+3)) =t ⇒x−3 =tx+3t ⇒(1−t)x =3t+3 ⇒x =((3t+3)/(1−t)) (dx/dx) =((3(1−t)+(3t+3))/((1−t)^2 )) =(6/((1−t)^2 )) x+3 =((3t+3)/(1−t)) +3 =((3t+3+3−3t)/(1−t)) =(6/(1−t)) ⇒A =∫_(1/4) ^1 (6/((1−t)^2 t^4 ((6/(1−t)))^8 ))dt =(1/6^7 ) ∫_(1/4) ^1 (((t−1)^8 )/((t−1)^2 t^4 ))dt =(1/6^7 ) ∫_(1/4) ^1 (((t−1)^6 )/t^4 ) dt =(1/6^7 ) ∫_(1/4) ^1 ((Σ_(k=0) ^6 C_6 ^(k ) t^k )/t^4 )dt =(1/6^7 ) Σ_(k=0) ^6 C_6 ^k ∫_(1/4) ^1 t^(k−4) dt =(1/6^7 ) Σ_(k=0and k≠3) ^6 C_6 ^k [(1/(k−3))t^(k−3) ]_(1/4) ^1 +(1/6^7 )C_6 ^3 [lnt]_(1/4) ^1 A =(1/6^7 ) Σ_(k=0 and k≠3) ^6 (C_6 ^k /(k−3)){1−(1/4^(k−3) )} +(C_6 ^3 /6^7 )×(2ln2)

$$\mathrm{A}\:=\int_{\mathrm{5}} ^{\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{4}} }\:\Rightarrow\mathrm{A}\:=\int_{\mathrm{5}} ^{\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{4}} \left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{4}} }\:=\int_{\mathrm{5}} ^{\infty} \:\frac{\mathrm{dx}}{\left(\frac{\mathrm{x}−\mathrm{3}}{\mathrm{x}+\mathrm{3}}\right)^{\mathrm{4}} \left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{8}} } \\ $$$$\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\frac{\mathrm{x}−\mathrm{3}}{\mathrm{x}+\mathrm{3}}\:=\mathrm{t}\:\Rightarrow\mathrm{x}−\mathrm{3}\:=\mathrm{tx}+\mathrm{3t}\:\Rightarrow\left(\mathrm{1}−\mathrm{t}\right)\mathrm{x}\:=\mathrm{3t}+\mathrm{3}\:\Rightarrow\mathrm{x}\:=\frac{\mathrm{3t}+\mathrm{3}}{\mathrm{1}−\mathrm{t}} \\ $$$$\frac{\mathrm{dx}}{\mathrm{dx}}\:=\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{t}\right)+\left(\mathrm{3t}+\mathrm{3}\right)}{\left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{2}} }\:=\frac{\mathrm{6}}{\left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{2}} } \\ $$$$\mathrm{x}+\mathrm{3}\:=\frac{\mathrm{3t}+\mathrm{3}}{\mathrm{1}−\mathrm{t}}\:+\mathrm{3}\:=\frac{\mathrm{3t}+\mathrm{3}+\mathrm{3}−\mathrm{3t}}{\mathrm{1}−\mathrm{t}}\:=\frac{\mathrm{6}}{\mathrm{1}−\mathrm{t}}\:\Rightarrow\mathrm{A}\:=\int_{\frac{\mathrm{1}}{\mathrm{4}}} ^{\mathrm{1}} \:\frac{\mathrm{6}}{\left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{2}} \mathrm{t}^{\mathrm{4}} \left(\frac{\mathrm{6}}{\mathrm{1}−\mathrm{t}}\right)^{\mathrm{8}} }\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{7}} }\:\int_{\frac{\mathrm{1}}{\mathrm{4}}} ^{\mathrm{1}} \:\:\frac{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{8}} }{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} \:\mathrm{t}^{\mathrm{4}} }\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{7}} }\:\int_{\frac{\mathrm{1}}{\mathrm{4}}} ^{\mathrm{1}} \:\:\frac{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{6}} }{\mathrm{t}^{\mathrm{4}} }\:\mathrm{dt}\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{7}} }\:\int_{\frac{\mathrm{1}}{\mathrm{4}}} ^{\mathrm{1}} \:\frac{\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{6}} \:\mathrm{C}_{\mathrm{6}} ^{\mathrm{k}\:} \:\mathrm{t}^{\mathrm{k}} }{\mathrm{t}^{\mathrm{4}} }\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{7}} }\:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{6}} \:\mathrm{C}_{\mathrm{6}} ^{\mathrm{k}} \:\int_{\frac{\mathrm{1}}{\mathrm{4}}} ^{\mathrm{1}} \:\mathrm{t}^{\mathrm{k}−\mathrm{4}} \:\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{7}} }\:\sum_{\mathrm{k}=\mathrm{0and}\:\mathrm{k}\neq\mathrm{3}} ^{\mathrm{6}} \:\mathrm{C}_{\mathrm{6}} ^{\mathrm{k}} \:\left[\frac{\mathrm{1}}{\mathrm{k}−\mathrm{3}}\mathrm{t}^{\mathrm{k}−\mathrm{3}} \right]_{\frac{\mathrm{1}}{\mathrm{4}}} ^{\mathrm{1}} \:\:\:+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{7}} }\mathrm{C}_{\mathrm{6}} ^{\mathrm{3}} \:\left[\mathrm{lnt}\right]_{\frac{\mathrm{1}}{\mathrm{4}}} ^{\mathrm{1}} \\ $$$$\mathrm{A}\:=\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{7}} }\:\sum_{\mathrm{k}=\mathrm{0}\:\mathrm{and}\:\mathrm{k}\neq\mathrm{3}} ^{\mathrm{6}} \:\frac{\mathrm{C}_{\mathrm{6}} ^{\mathrm{k}} }{\mathrm{k}−\mathrm{3}}\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{k}−\mathrm{3}} }\right\}\:+\frac{\mathrm{C}_{\mathrm{6}} ^{\mathrm{3}} }{\mathrm{6}^{\mathrm{7}} }×\left(\mathrm{2ln2}\right) \\ $$

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