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Question Number 63615 by aliesam last updated on 06/Jul/19

Commented by mathmax by abdo last updated on 06/Jul/19
$changement x=−t give ∫_(−1) ^1 xln(1+10^x )dx =−∫_(−1) ^1 (−t)ln(1+10^(−t) )(−dt) =−∫_(−1) ^1 t ln(1+(1/(10^t )))dt =−∫_(−1) ^1 t {ln(1+10^t )−ln(10^t )}dt =−∫_(−1) ^1 tln(1+10^t ) +∫_(−1) ^1 tln(10^t )dt ⇒ 2∫_(−1) ^1 xln(1+10^x )dx = ∫_(−1) ^1 t^2 ln(10)dt =2ln(10) ∫_0 ^1 t^2 dt =2ln(10)×(1/3) =(2/3)ln(10) .$
$c h a n g e m e n t x = - t g i v e \int_{- 1}^{1} x l n (1 + 10^{x}) d x = - \int_{- 1}^{1} (- t) l n (1 + 10^{- t}) (- d t)$ $= - \int_{- 1}^{1} t l n (1 + \frac{1}{10^{t}}) d t = - \int_{- 1}^{1} t {l n (1 + 10^{t}) - l n (10^{t})} d t$ $= - \int_{- 1}^{1} t l n (1 + 10^{t}) + \int_{- 1}^{1} t l n (10^{t}) d t \Rightarrow$ $2 \int_{- 1}^{1} x l n (1 + 10^{x}) d x = \int_{- 1}^{1} t^{2} l n (10) d t = 2 l n (10) \int_{0}^{1} t^{2} d t = 2 l n (10) \times \frac{1}{3}$ $= \frac{2}{3} l n (10) .$
Commented by aliesam last updated on 06/Jul/19

$g o d b l e s s y o u s i r$
Commented by mathmax by abdo last updated on 06/Jul/19

$y o u a r e w e l c o m e .$
Commented by MJS last updated on 06/Jul/19

$great! but of course the final answer is \frac{1}{3} \ln 10$
Commented by Prithwish sen last updated on 06/Jul/19

$excellent sir .$
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