Question Number 116245 by mathmax by abdo last updated on 02/Oct/20
$$\mathrm{calculate}\:\:\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{2x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{5}} } \\ $$
Answered by MJS_new last updated on 02/Oct/20
![∫(dx/((2x^2 −1)^5 ))= [Ostrogradski] =((x(840x^6 −1540x^4 +1022x^2 −279))/(384(2x^2 −1)^4 ))+((35)/(128))∫(dx/(2x^2 −1))= =((x(840x^6 −1540x^4 +1022x^2 −279))/(384(2x^2 −1)^4 ))+((35(√2))/(512))ln (((√2)x−1)/( (√2)x+1)) +C ⇒ ∫_1 ^∞ (dx/((2x^2 −1)^5 ))=−((43)/(384))+((35(√2))/(256))ln (1+(√2))](https://www.tinkutara.com/question/Q116255.png)
$$\int\frac{{dx}}{\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{5}} }= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}\right] \\ $$$$=\frac{{x}\left(\mathrm{840}{x}^{\mathrm{6}} −\mathrm{1540}{x}^{\mathrm{4}} +\mathrm{1022}{x}^{\mathrm{2}} −\mathrm{279}\right)}{\mathrm{384}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{4}} }+\frac{\mathrm{35}}{\mathrm{128}}\int\frac{{dx}}{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}= \\ $$$$=\frac{{x}\left(\mathrm{840}{x}^{\mathrm{6}} −\mathrm{1540}{x}^{\mathrm{4}} +\mathrm{1022}{x}^{\mathrm{2}} −\mathrm{279}\right)}{\mathrm{384}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{4}} }+\frac{\mathrm{35}\sqrt{\mathrm{2}}}{\mathrm{512}}\mathrm{ln}\:\frac{\sqrt{\mathrm{2}}{x}−\mathrm{1}}{\:\sqrt{\mathrm{2}}{x}+\mathrm{1}}\:+{C} \\ $$$$\Rightarrow \\ $$$$\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{{dx}}{\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{5}} }=−\frac{\mathrm{43}}{\mathrm{384}}+\frac{\mathrm{35}\sqrt{\mathrm{2}}}{\mathrm{256}}\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$
Commented by Bird last updated on 02/Oct/20
$${thank}\:{you}\:{sir} \\ $$
Answered by 1549442205PVT last updated on 03/Oct/20
![(2/(2x^2 −1))=(1/( (√2)x−1))−(1/( (√2)x+1)).Hence, Put (1/( (√2)x−1))=a,(1/( (√2)x+1))=b⇒2ab=a−b (1/((2x^2 −1)^5 ))=(1/(32))×((2/(2x^2 −1)))^5 =(1/(32))(a−b)^5 We have (a−b)^5 =a^5 −b^5 −5ab(a^3 −b^3 )+10(ab)^2 (a−b) a^5 −b^5 −(5/2)(a−b)(a^3 −b^3 )+((10)/4)(a−b)^3 =a^5 −b^5 −(5/2)(a^4 +b^4 )+(5/2)ab(a^2 +b^2 )+(5/2)[a^3 −b^3 −3ab(a−b)] =a^5 −b^5 −(5/2)(a^4 +b^4 )+(5/4)(a−b)(a^2 +b^2 )+(5/2)[a^3 −b^3 −3ab(a−b)]] =a^5 −b^5 −(5/2)(a^4 +b^4 )+(5/4)(a^3 −b^3 )−(5/4)ab(a−b)+(5/2)[a^3 −b^3 −3ab(a−b)]] =a^5 −b^5 −(5/2)(a^4 +b^4 )+((15)/4)(a^3 −b^3 )−((35)/8)(a−b)^2 =a^5 −b^5 −(5/2)(a^4 +b^4 )+((15)/4)(a^3 −b^3 )−((35)/8)(a^2 +b^2 )+((35)/8)(a−b) I=∫_1 ^∞ [(1/(32((√2)x−1)^5 ))−(1/(32((√2)x+1)^5 ))−(5/(64((√2)x−1)^4 )) −(5/(64((√2)x+1)^4 ))+((15)/(128((√2)x−1)^3 ))−((15)/(128((√2)x+1)^3 )) −((35)/(256((√2)x−1)^2 ))−((35)/(256((√2)x+1)^2 ))+((35)/(256))(a−b)]dx Put (√2) x−1=u,(√2)x+1=v⇒(√2)dx=du=dv.Hence, I=−(1/(128(√2)((√2)x−1)^4 ))+(1/(128(√2)((√2)x+1)^4 )) +(5/(192(√2)((√2)x−1)^3 ))+(5/(192(√2)((√2)x+1)^3 )) −((15)/( 256(√2)((√2)x−1)^2 ))+((15)/( 256(√2)((√2)x+1)^2 )) +((35)/( 256(√2)((√2)x−1)))+((35)/( 256(√2)((√2)x+1)))]_1 ^∞ +((35)/(256(√2)))ln∣(((√2)x−1)/( (√2)x+1))∣ =[((−(16(√2)x^3 +8(√2)x))/(128(√2)(2x^2 −1)^4 ))+((5(4(√2)x^3 +6(√2)x))/(192(√2)(2x^2 −1)^3 )) −((15×(4(√2)x))/( 256(√2)(2x^2 −1)^2 ))+((35×2(√2)x)/( 256(√2)(2x^2 −1)))+((35)/(256(√2)))ln∣(((√2)x−1)/( (√2)x+1))∣]_1 ^∞ =−(((−24(√2))/(128(√2)))+((50(√2))/( 192(√2)))−((60(√2))/( 256(√2)))+((70(√2))/( 256(√2)))+((35)/(256))ln(((√2)−1)/( (√2)+1))) =−((43)/(384))−((35(√2))/(256))ln((√2)−1)≈0.058434](https://www.tinkutara.com/question/Q116269.png)
$$\frac{\mathrm{2}}{\mathrm{2x}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\mathrm{x}−\mathrm{1}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}}.\mathrm{Hence}, \\ $$$$\mathrm{Put}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\mathrm{x}−\mathrm{1}}=\mathrm{a},\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}}=\mathrm{b}\Rightarrow\mathrm{2ab}=\mathrm{a}−\mathrm{b} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{2x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{5}} }=\frac{\mathrm{1}}{\mathrm{32}}×\left(\frac{\mathrm{2}}{\mathrm{2x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{32}}\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{5}} \\ $$$$\mathrm{We}\:\mathrm{have}\:\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{5}} \\ $$$$=\mathrm{a}^{\mathrm{5}} −\mathrm{b}^{\mathrm{5}} −\mathrm{5ab}\left(\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} \right)+\mathrm{10}\left(\mathrm{ab}\right)^{\mathrm{2}} \left(\mathrm{a}−\mathrm{b}\right) \\ $$$$\mathrm{a}^{\mathrm{5}} −\mathrm{b}^{\mathrm{5}} −\frac{\mathrm{5}}{\mathrm{2}}\left(\mathrm{a}−\mathrm{b}\right)\left(\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} \right)+\frac{\mathrm{10}}{\mathrm{4}}\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{3}} \\ $$$$=\mathrm{a}^{\mathrm{5}} −\mathrm{b}^{\mathrm{5}} −\frac{\mathrm{5}}{\mathrm{2}}\left(\mathrm{a}^{\mathrm{4}} +\mathrm{b}^{\mathrm{4}} \right)+\frac{\mathrm{5}}{\mathrm{2}}\mathrm{ab}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)+\frac{\mathrm{5}}{\mathrm{2}}\left[\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} −\mathrm{3ab}\left(\mathrm{a}−\mathrm{b}\right)\right]\: \\ $$$$\left.=\mathrm{a}^{\mathrm{5}} −\mathrm{b}^{\mathrm{5}} −\frac{\mathrm{5}}{\mathrm{2}}\left(\mathrm{a}^{\mathrm{4}} +\mathrm{b}^{\mathrm{4}} \right)+\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{a}−\mathrm{b}\right)\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)+\frac{\mathrm{5}}{\mathrm{2}}\left[\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} −\mathrm{3ab}\left(\mathrm{a}−\mathrm{b}\right)\right]\right]\: \\ $$$$\left.=\mathrm{a}^{\mathrm{5}} −\mathrm{b}^{\mathrm{5}} −\frac{\mathrm{5}}{\mathrm{2}}\left(\mathrm{a}^{\mathrm{4}} +\mathrm{b}^{\mathrm{4}} \right)+\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} \right)−\frac{\mathrm{5}}{\mathrm{4}}\mathrm{ab}\left(\mathrm{a}−\mathrm{b}\right)+\frac{\mathrm{5}}{\mathrm{2}}\left[\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} −\mathrm{3ab}\left(\mathrm{a}−\mathrm{b}\right)\right]\right]\: \\ $$$$=\mathrm{a}^{\mathrm{5}} −\mathrm{b}^{\mathrm{5}} −\frac{\mathrm{5}}{\mathrm{2}}\left(\mathrm{a}^{\mathrm{4}} +\mathrm{b}^{\mathrm{4}} \right)+\frac{\mathrm{15}}{\mathrm{4}}\left(\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} \right)−\frac{\mathrm{35}}{\mathrm{8}}\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} \\ $$$$=\mathrm{a}^{\mathrm{5}} −\mathrm{b}^{\mathrm{5}} −\frac{\mathrm{5}}{\mathrm{2}}\left(\mathrm{a}^{\mathrm{4}} +\mathrm{b}^{\mathrm{4}} \right)+\frac{\mathrm{15}}{\mathrm{4}}\left(\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} \right)−\frac{\mathrm{35}}{\mathrm{8}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)+\frac{\mathrm{35}}{\mathrm{8}}\left(\mathrm{a}−\mathrm{b}\right) \\ $$$$\mathrm{I}=\int_{\mathrm{1}} ^{\infty} \left[\frac{\mathrm{1}}{\mathrm{32}\left(\sqrt{\mathrm{2}}\mathrm{x}−\mathrm{1}\right)^{\mathrm{5}} }−\frac{\mathrm{1}}{\mathrm{32}\left(\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}\right)^{\mathrm{5}} }−\frac{\mathrm{5}}{\mathrm{64}\left(\sqrt{\mathrm{2}}\mathrm{x}−\mathrm{1}\right)^{\mathrm{4}} }\right. \\ $$$$−\frac{\mathrm{5}}{\mathrm{64}\left(\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}\right)^{\mathrm{4}} }+\frac{\mathrm{15}}{\mathrm{128}\left(\sqrt{\mathrm{2}}\mathrm{x}−\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{15}}{\mathrm{128}\left(\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\left.−\frac{\mathrm{35}}{\mathrm{256}\left(\sqrt{\mathrm{2}}\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{35}}{\mathrm{256}\left(\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{35}}{\mathrm{256}}\left(\mathrm{a}−\mathrm{b}\right)\right]\mathrm{dx} \\ $$$$\mathrm{Put}\:\sqrt{\mathrm{2}}\:\mathrm{x}−\mathrm{1}=\mathrm{u},\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}=\mathrm{v}\Rightarrow\sqrt{\mathrm{2}}\mathrm{dx}=\mathrm{du}=\mathrm{dv}.\mathrm{Hence}, \\ $$$$\mathrm{I}=−\frac{\mathrm{1}}{\mathrm{128}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}\mathrm{x}−\mathrm{1}\right)^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{128}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$+\frac{\mathrm{5}}{\mathrm{192}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}\mathrm{x}−\mathrm{1}\right)^{\mathrm{3}} }+\frac{\mathrm{5}}{\mathrm{192}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$−\frac{\mathrm{15}}{\:\mathrm{256}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{15}}{\:\mathrm{256}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\left.+\frac{\mathrm{35}}{\:\mathrm{256}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}\mathrm{x}−\mathrm{1}\right)}+\frac{\mathrm{35}}{\:\mathrm{256}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}\right)}\right]_{\mathrm{1}} ^{\infty} \\ $$$$+\frac{\mathrm{35}}{\mathrm{256}\sqrt{\mathrm{2}}}\mathrm{ln}\mid\frac{\sqrt{\mathrm{2}}\mathrm{x}−\mathrm{1}}{\:\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}}\mid \\ $$$$=\left[\frac{−\left(\mathrm{16}\sqrt{\mathrm{2}}\mathrm{x}^{\mathrm{3}} +\mathrm{8}\sqrt{\mathrm{2}}\mathrm{x}\right)}{\mathrm{128}\sqrt{\mathrm{2}}\left(\mathrm{2x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{4}} }+\frac{\mathrm{5}\left(\mathrm{4}\sqrt{\mathrm{2}}\mathrm{x}^{\mathrm{3}} +\mathrm{6}\sqrt{\mathrm{2}}\mathrm{x}\right)}{\mathrm{192}\sqrt{\mathrm{2}}\left(\mathrm{2x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} }\right. \\ $$$$\left.−\frac{\mathrm{15}×\left(\mathrm{4}\sqrt{\mathrm{2}}\mathrm{x}\right)}{\:\mathrm{256}\sqrt{\mathrm{2}}\left(\mathrm{2x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{35}×\mathrm{2}\sqrt{\mathrm{2}}\mathrm{x}}{\:\mathrm{256}\sqrt{\mathrm{2}}\left(\mathrm{2x}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\mathrm{35}}{\mathrm{256}\sqrt{\mathrm{2}}}\mathrm{ln}\mid\frac{\sqrt{\mathrm{2}}\mathrm{x}−\mathrm{1}}{\:\sqrt{\mathrm{2}}\mathrm{x}+\mathrm{1}}\mid\right]_{\mathrm{1}} ^{\infty} \\ $$$$=−\left(\frac{−\mathrm{24}\sqrt{\mathrm{2}}}{\mathrm{128}\sqrt{\mathrm{2}}}+\frac{\mathrm{50}\sqrt{\mathrm{2}}}{\:\mathrm{192}\sqrt{\mathrm{2}}}−\frac{\mathrm{60}\sqrt{\mathrm{2}}}{\:\mathrm{256}\sqrt{\mathrm{2}}}+\frac{\mathrm{70}\sqrt{\mathrm{2}}}{\:\mathrm{256}\sqrt{\mathrm{2}}}+\frac{\mathrm{35}}{\mathrm{256}}\mathrm{ln}\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}\right) \\ $$$$=−\frac{\mathrm{43}}{\mathrm{384}}−\frac{\mathrm{35}\sqrt{\mathrm{2}}}{\mathrm{256}}\mathrm{ln}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\approx\mathrm{0}.\mathrm{058434} \\ $$
Commented by Bird last updated on 02/Oct/20
$${thank}\:{you}\:{sir}\:{for}\:{this}\:{hardwork} \\ $$
Commented by 1549442205PVT last updated on 03/Oct/20
$$\mathrm{Thank}\:\mathrm{Sir}.\mathrm{You}\:\mathrm{are}\:\mathrm{welcome}. \\ $$
Answered by Bird last updated on 03/Oct/20
![I =∫_1 ^∞ (dx/((2x^2 −1)^5 )) ⇒ I =∫_1 ^∞ (dx/((x(√2)−1)^5 (x(√2)+1)^5 )) =∫_1 ^∞ (dx/((((x(√2)−1)/(x(√2)+1)))^5 (x(√2)+1)^(10) )) we do the changement ((x(√2)−1)/(x(√2)+1))=t ⇒x(√2)−1 =tx(√2)+t ⇒ ((√2)−t(√2))x =t+1 ⇒x =((t+1)/( (√2)(1−t))) ⇒(dx/dt) =(1/( (√2))).((1−t−(t+1)(−1))/((1−t)^2 )) =(1/( (√2))).(2/((t−1)^2 )) also x(√2)+1 =((t+1)/(1−t))+1 =((t+1+1−t)/(1−t)) =(2/(1−t)) ⇒ ⇒I =∫_(((√2)−1)/( (√2)+1)) ^1 (1/(t^5 ((2/(1−t)))^(10) ))×((√2)/((t−1)^2 ))dt =((√2)/2^(10) )∫_(3−2(√2)) ^1 (((t−1)^8 )/t^5 )dt =((√2)/2^(10) ) ∫_(3−2(√2)) ^1 ((Σ_(k=0) ^8 C_8 ^k t^k (−1)^(8−k) )/t^5 )dt =((√2)/2^(10) ) Σ_(k=0) ^8 (−1)^k C_8 ^k ∫_(3−2(√2)) ^1 t^(k−5) dt =((√2)/2^(10) )Σ_(k=0 and k≠4) ^8 (−1)^k C_8 ^k [(1/(k−4))t^(k−4) ]_(3−2(√2)) ^1 +((√2)/2^(10) ) C_8 ^4 [lnt]_(3−2(√2)) ^1 I=((√2)/2^(10) ) Σ_(k=0and k≠4) ^8 (((−1)^(k ) C_8 ^k )/(k−4)){1−(3−2(√2))^(k−4) } −((√2)/2^(10) ) C_8 ^k ln(3−2(√2))](https://www.tinkutara.com/question/Q116328.png)
$${I}\:=\int_{\mathrm{1}} ^{\infty} \:\frac{{dx}}{\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{5}} }\:\Rightarrow \\ $$$${I}\:=\int_{\mathrm{1}} ^{\infty} \:\frac{{dx}}{\left({x}\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{5}} \left({x}\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{5}} } \\ $$$$=\int_{\mathrm{1}} ^{\infty} \:\frac{{dx}}{\left(\frac{{x}\sqrt{\mathrm{2}}−\mathrm{1}}{{x}\sqrt{\mathrm{2}}+\mathrm{1}}\right)^{\mathrm{5}} \left({x}\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{10}} } \\ $$$${we}\:{do}\:{the}\:{changement}\:\frac{{x}\sqrt{\mathrm{2}}−\mathrm{1}}{{x}\sqrt{\mathrm{2}}+\mathrm{1}}={t} \\ $$$$\Rightarrow{x}\sqrt{\mathrm{2}}−\mathrm{1}\:={tx}\sqrt{\mathrm{2}}+{t}\:\Rightarrow \\ $$$$\left(\sqrt{\mathrm{2}}−{t}\sqrt{\mathrm{2}}\right){x}\:={t}+\mathrm{1}\:\Rightarrow{x}\:=\frac{{t}+\mathrm{1}}{\:\sqrt{\mathrm{2}}\left(\mathrm{1}−{t}\right)} \\ $$$$\Rightarrow\frac{{dx}}{{dt}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}.\frac{\mathrm{1}−{t}−\left({t}+\mathrm{1}\right)\left(−\mathrm{1}\right)}{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}.\frac{\mathrm{2}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:{also}\:{x}\sqrt{\mathrm{2}}+\mathrm{1}\:=\frac{{t}+\mathrm{1}}{\mathrm{1}−{t}}+\mathrm{1} \\ $$$$=\frac{{t}+\mathrm{1}+\mathrm{1}−{t}}{\mathrm{1}−{t}}\:=\frac{\mathrm{2}}{\mathrm{1}−{t}}\:\Rightarrow \\ $$$$\Rightarrow{I}\:=\int_{\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{{t}^{\mathrm{5}} \left(\frac{\mathrm{2}}{\mathrm{1}−{t}}\right)^{\mathrm{10}} }×\frac{\sqrt{\mathrm{2}}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}^{\mathrm{10}} }\int_{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} ^{\mathrm{1}} \:\:\:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{8}} }{{t}^{\mathrm{5}} }{dt} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}^{\mathrm{10}} }\:\int_{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} ^{\mathrm{1}} \frac{\sum_{{k}=\mathrm{0}} ^{\mathrm{8}} \:{C}_{\mathrm{8}} ^{{k}} \:{t}^{{k}} \left(−\mathrm{1}\right)^{\mathrm{8}−{k}} }{{t}^{\mathrm{5}} }{dt} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}^{\mathrm{10}} }\:\sum_{{k}=\mathrm{0}} ^{\mathrm{8}} \:\left(−\mathrm{1}\right)^{{k}} \:{C}_{\mathrm{8}} ^{{k}} \:\int_{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} ^{\mathrm{1}} {t}^{{k}−\mathrm{5}} {dt} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}^{\mathrm{10}} }\sum_{{k}=\mathrm{0}\:{and}\:{k}\neq\mathrm{4}} ^{\mathrm{8}} \left(−\mathrm{1}\right)^{{k}} \:{C}_{\mathrm{8}} ^{{k}} \left[\frac{\mathrm{1}}{{k}−\mathrm{4}}{t}^{{k}−\mathrm{4}} \right]_{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} ^{\mathrm{1}} \\ $$$$+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}^{\mathrm{10}} }\:{C}_{\mathrm{8}} ^{\mathrm{4}} \:\left[{lnt}\right]_{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} ^{\mathrm{1}} \\ $$$${I}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}^{\mathrm{10}} }\:\sum_{{k}=\mathrm{0}{and}\:{k}\neq\mathrm{4}} ^{\mathrm{8}} \:\frac{\left(−\mathrm{1}\right)^{{k}\:} {C}_{\mathrm{8}} ^{{k}} }{{k}−\mathrm{4}}\left\{\mathrm{1}−\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)^{{k}−\mathrm{4}} \right\} \\ $$$$−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}^{\mathrm{10}} }\:{C}_{\mathrm{8}} ^{{k}} {ln}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$