∫_0 ^1 ln(1+(1/(n^2 x^2 )))dx


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Question Number 99818 by Ar Brandon last updated on 23/Jun/20

$\int_{0}^{1} \ln (1 + \frac{1}{n^{2} x^{2}}) dx$
	Answered by smridha last updated on 23/Jun/20

	$s i m p l y u s i n g b y p a r t s y o u g e t$ $A n s : l n (1 + \frac{1}{n^{2}}) + \frac{2}{n} \tan^{- 1} (n)$
	Commented by Ar Brandon last updated on 23/Jun/20

	$THANKS$
	Answered by mathmax by abdo last updated on 23/Jun/20
	$changement (1/(nx)) =t give nx =(1/t) ⇒x =(1/(nt)) ⇒dx =−(1/(nt^2 ))dt ⇒ I=∫_0 ^1 ln(1+(1/(n^2 t^2 )))dx =∫_(1/n) ^(+∞) ln(1+t^2 )(dt/(nt^2 )) =(1/n) ∫_(1/n) ^(+∞) ((ln(1+t^2 ))/t^2 ) dt by parts ∫_(1/n) ^(+∞) ((ln(1+t^2 ))/t^2 )dt =[−(1/t)ln(1+t^2 )]_(1/n) ^(+∞) +∫_(1/n) ^(+∞) (1/t)×((2t)/(1+t^2 ))dt =nln(1+(1/n^2 )) +2 [arctant]_(1/n) ^(+∞) =n ln(1+(1/n^2 ))+2{(π/2) −arctan((1/n))} =n ln(1+(1/n^2 )) +2arctan(n) ⇒I =ln(1+(1/n^2 ))+((2arctan(n))/n)$
	$changement \frac{1}{nx} = t give nx = \frac{1}{t} \Rightarrow x = \frac{1}{nt} \Rightarrow dx = - \frac{1}{{nt}^{2}} dt \Rightarrow$ $I = \int_{0}^{1} \ln (1 + \frac{1}{n^{2} t^{2}}) dx = \int_{\frac{1}{n}}^{+ \infty} \ln (1 + t^{2}) \frac{dt}{{nt}^{2}} = \frac{1}{n} \int_{\frac{1}{n}}^{+ \infty} \frac{\ln (1 + t^{2})}{t^{2}} dt by parts$ $\int_{\frac{1}{n}}^{+ \infty} \frac{\ln (1 + t^{2})}{t^{2}} dt = {[- \frac{1}{t} \ln (1 + t^{2})]}_{\frac{1}{n}}^{+ \infty} + \int_{\frac{1}{n}}^{+ \infty} \frac{1}{t} \times \frac{2 t}{1 + t^{2}} dt$ $= nln (1 + \frac{1}{n^{2}}) + 2 {[arctant]}_{\frac{1}{n}}^{+ \infty} = n \ln (1 + \frac{1}{n^{2}}) + 2 {\frac{π}{2} - \arctan (\frac{1}{n})}$ $= n \ln (1 + \frac{1}{n^{2}}) + 2 \arctan (n) \Rightarrow I = \ln (1 + \frac{1}{n^{2}}) + \frac{2 \arctan (n)}{n}$
	Commented by Ar Brandon last updated on 23/Jun/20
	cool, thanks
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