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Question-63615




Question Number 63615 by aliesam last updated on 06/Jul/19
Commented by mathmax by abdo last updated on 06/Jul/19
changement  x=−t  give ∫_(−1) ^1  xln(1+10^x )dx =−∫_(−1) ^1  (−t)ln(1+10^(−t) )(−dt)  =−∫_(−1) ^1  t ln(1+(1/(10^t )))dt =−∫_(−1) ^1  t {ln(1+10^t )−ln(10^t )}dt  =−∫_(−1) ^1  tln(1+10^t ) +∫_(−1) ^1 tln(10^t )dt ⇒                2∫_(−1) ^1  xln(1+10^x )dx = ∫_(−1) ^1 t^2 ln(10)dt =2ln(10) ∫_0 ^1  t^2 dt =2ln(10)×(1/3)  =(2/3)ln(10) .
$${changement}\:\:{x}=−{t}\:\:{give}\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:{xln}\left(\mathrm{1}+\mathrm{10}^{{x}} \right){dx}\:=−\int_{−\mathrm{1}} ^{\mathrm{1}} \:\left(−{t}\right){ln}\left(\mathrm{1}+\mathrm{10}^{−{t}} \right)\left(−{dt}\right) \\ $$$$=−\int_{−\mathrm{1}} ^{\mathrm{1}} \:{t}\:{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{10}^{{t}} }\right){dt}\:=−\int_{−\mathrm{1}} ^{\mathrm{1}} \:{t}\:\left\{{ln}\left(\mathrm{1}+\mathrm{10}^{{t}} \right)−{ln}\left(\mathrm{10}^{{t}} \right)\right\}{dt} \\ $$$$=−\int_{−\mathrm{1}} ^{\mathrm{1}} \:{tln}\left(\mathrm{1}+\mathrm{10}^{{t}} \right)\:+\int_{−\mathrm{1}} ^{\mathrm{1}} {tln}\left(\mathrm{10}^{{t}} \right){dt}\:\Rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\mathrm{2}\int_{−\mathrm{1}} ^{\mathrm{1}} \:{xln}\left(\mathrm{1}+\mathrm{10}^{{x}} \right){dx}\:=\:\int_{−\mathrm{1}} ^{\mathrm{1}} {t}^{\mathrm{2}} {ln}\left(\mathrm{10}\right){dt}\:=\mathrm{2}{ln}\left(\mathrm{10}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}} {dt}\:=\mathrm{2}{ln}\left(\mathrm{10}\right)×\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}{ln}\left(\mathrm{10}\right)\:. \\ $$
Commented by aliesam last updated on 06/Jul/19
god bless you sir
$${god}\:{bless}\:{you}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 06/Jul/19
you are welcome.
$${you}\:{are}\:{welcome}. \\ $$
Commented by MJS last updated on 06/Jul/19
great! but of course the final answer is (1/3)ln 10
$$\mathrm{great}!\:\mathrm{but}\:\mathrm{of}\:\mathrm{course}\:\mathrm{the}\:\mathrm{final}\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\mathrm{10} \\ $$
Commented by Prithwish sen last updated on 06/Jul/19
excellent sir.
$$\mathrm{excellent}\:\mathrm{sir}. \\ $$

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