Question-85391

Question Number 85391 by sakeefhasan05@gmail.com last updated on 21/Mar/20

Commented by sakeefhasan05@gmail.com last updated on 21/Mar/20

I have got ((8/(63))) as answer whoever check it pls

$$\mathrm{I}\:\mathrm{have}\:\mathrm{got}\:\left(\frac{\mathrm{8}}{\mathrm{63}}\right)\:\mathrm{as}\:\mathrm{answer}\: \\ $$$$\mathrm{whoever}\:\mathrm{check}\:\mathrm{it}\:\mathrm{pls}\: \\ $$

Commented by mathmax by abdo last updated on 21/Mar/20

I =∫_0 ^∞ (dx/((x+(√(1+x^2 )))^8 )) changement x =sh(t) give I =∫_0 ^∞ ((ch(t))/((sht +ch(t))^8 ))dt =(1/2)∫_0 ^∞ ((e^t +e^(−t) )/e^(8t) )dt =(1/2)∫_0 ^∞ ( e^(−7t) +e^(−9t) )dt =−(1/2)[ (1/7)e^((7t) +(1/9)e^(−9t) ]_0 ^(+∞) =−(1/2)(−(1/7)−(1/9)) =(1/2)((1/7)+(1/9)) =(1/2)×((16)/(63)) =(8/(63))

$${I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{8}} }\:{changement}\:{x}\:={sh}\left({t}\right)\:{give} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ch}\left({t}\right)}{\left({sht}\:+{ch}\left({t}\right)\right)^{\mathrm{8}} }{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{{t}} \:+{e}^{−{t}} }{{e}^{\mathrm{8}{t}} }{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \left(\:{e}^{−\mathrm{7}{t}} \:+{e}^{−\mathrm{9}{t}} \right){dt}\:=−\frac{\mathrm{1}}{\mathrm{2}}\left[\:\frac{\mathrm{1}}{\mathrm{7}}{e}^{\left(\mathrm{7}{t}\right.} \:+\frac{\mathrm{1}}{\mathrm{9}}{e}^{−\mathrm{9}{t}} \right]_{\mathrm{0}} ^{+\infty} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{7}}−\frac{\mathrm{1}}{\mathrm{9}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{9}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{16}}{\mathrm{63}}\:=\frac{\mathrm{8}}{\mathrm{63}} \\ $$

Commented by MJS last updated on 21/Mar/20

$$\mathrm{great},\:\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{think}\:\mathrm{of}\:\mathrm{this}\:\mathrm{path}… \\ $$

Commented by sakeefhasan05@gmail.com last updated on 21/Mar/20

$$\mathrm{thank}\:\mathrm{u}\:\mathrm{very}\:\mathrm{much} \\ $$

Commented by mathmax by abdo last updated on 21/Mar/20

$${thank}\:{you}\:{sir}. \\ $$

Answered by MJS last updated on 21/Mar/20

F(x)=∫(dx/((x+(√(1+x^2 )))^8 ))=∫(x−(√(1+x^2 )))^8 dx= =∫(128x^8 +256x^6 +160x^4 +32x^2 +1)d∫x− −8∫x(√(x^2 +1))(16x^6 +24x^4 +10x^2 +1)dx= =((128)/9)x^9 +((256)/7)x^7 +32x^5 +((32)/3)x^3 +x− −8(√((x^2 +1)^3 ))(((16)/9)x^6 +((40)/(21))x^4 +((10)/(21))x^2 +(1/(63))) +C lim_(x→∞) F(x) =0 (? please check!) ⇒ ∫_0 ^∞ (dx/((x+(√(1+x^2 )))^8 ))=(8/(63))

$${F}\left({x}\right)=\int\frac{{dx}}{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{8}} }=\int\left({x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{8}} {dx}= \\ $$$$=\int\left(\mathrm{128}{x}^{\mathrm{8}} +\mathrm{256}{x}^{\mathrm{6}} +\mathrm{160}{x}^{\mathrm{4}} +\mathrm{32}{x}^{\mathrm{2}} +\mathrm{1}\right){d}\int{x}− \\ $$$$\:\:\:\:\:−\mathrm{8}\int{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\left(\mathrm{16}{x}^{\mathrm{6}} +\mathrm{24}{x}^{\mathrm{4}} +\mathrm{10}{x}^{\mathrm{2}} +\mathrm{1}\right){dx}= \\ $$$$=\frac{\mathrm{128}}{\mathrm{9}}{x}^{\mathrm{9}} +\frac{\mathrm{256}}{\mathrm{7}}{x}^{\mathrm{7}} +\mathrm{32}{x}^{\mathrm{5}} +\frac{\mathrm{32}}{\mathrm{3}}{x}^{\mathrm{3}} +{x}− \\ $$$$\:\:\:\:\:−\mathrm{8}\sqrt{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\left(\frac{\mathrm{16}}{\mathrm{9}}{x}^{\mathrm{6}} +\frac{\mathrm{40}}{\mathrm{21}}{x}^{\mathrm{4}} +\frac{\mathrm{10}}{\mathrm{21}}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{63}}\right)\:+{C} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{F}\left({x}\right)\:=\mathrm{0}\:\left(?\:\mathrm{please}\:\mathrm{check}!\right)\:\Rightarrow\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{8}} }=\frac{\mathrm{8}}{\mathrm{63}} \\ $$

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