let A_n = ∫_(1/n) ^n (1+(1/x^2 ))arctanx dx 1) calculate A_n 2) find lim_(n→+∞) A


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Question Number 35049 by math khazana by abdo last updated on 14/May/18

$l e t A_{n} = \int_{\frac{1}{n}}^{n} (1 + \frac{1}{x^{2}}) a r c t a n x d x$ $1) c a l c u l a t e A_{n}$ $2) f i n d {l i m}_{n \to + \infty} A_{n}$
Commented by abdo mathsup 649 cc last updated on 16/May/18
$1) let integrate by parts u^′ =1+(1/x^2 ) and v=arctanx A_n = [(1−(1/x))arctanx]_(1/n) ^n −∫_(1/n) ^n (1−(1/x)) (dx/(1+x^2 )) =(1−(1/n))arctan(n) −(1−n)arctan((1/n)) − ∫_(1/n) ^n (dx/(1+x^2 )) + ∫_(1/n) ^n (dx/(x(1+x^2 ))) but ∫_(1/n) ^n (dx/(1+x^2 )) = arctan(n) −arctan((1/n)) ∫_(1/n) ^n (1/(x(1+x^2 ))) dx= ∫_(1/n) ^n ((1/x) −(x/(1+x^2 )))dx =ln(n) −ln((1/n)) −(1/2)[ln(1+x^2 )]_(1/n) ^n = 2ln(n) −(1/2){ ln(1+n^2 ) −ln(1+(1/n^2 ))}⇒ A_n = (1−(1/n))arctan(n)−(1−n)arctan((1/n)) −arctan(n) +arctan((1/n)) +2ln(n)−(1/2)(2ln(n)) A_n =−((arctan(n))/n) +n arctan((1/n)) +(π/2) −2arctan(n) +ln(n) 2) we have lim_(n→+∞) ( −((arctan(n))/n) +n arctan((1/n)) +(π/2) −2arctsn(n)) =0 +1 −(π/2) −π but lim_(n→+∞) ln(n) =+∞ ⇒ lim_(n→+∞) A_n = +∞.$
$1) l e t i n t e g r a t e b y p a r t s u^{^{'}} = 1 + \frac{1}{x^{2}} a n d v = a r c t a n x$ $A_{n} = {[(1 - \frac{1}{x}) a r c t a n x]}_{\frac{1}{n}}^{n} - \int_{\frac{1}{n}}^{n} (1 - \frac{1}{x}) \frac{d x}{1 + x^{2}}$ $= (1 - \frac{1}{n}) a r c t a n (n) - (1 - n) a r c t a n (\frac{1}{n})$ $- \int_{\frac{1}{n}}^{n} \frac{d x}{1 + x^{2}} + \int_{\frac{1}{n}}^{n} \frac{d x}{x (1 + x^{2})} b u t$ $\int_{\frac{1}{n}}^{n} \frac{d x}{1 + x^{2}} = a r c t a n (n) - a r c t a n (\frac{1}{n})$ $\int_{\frac{1}{n}}^{n} \frac{1}{x (1 + x^{2})} d x = \int_{\frac{1}{n}}^{n} (\frac{1}{x} - \frac{x}{1 + x^{2}}) d x$ $= l n (n) - l n (\frac{1}{n}) - \frac{1}{2} {[l n (1 + x^{2})]}_{\frac{1}{n}}^{n}$ $= 2 l n (n) - \frac{1}{2} {l n (1 + n^{2}) - l n (1 + \frac{1}{n^{2}})} \Rightarrow$ $A_{n} = (1 - \frac{1}{n}) a r c t a n (n) - (1 - n) a r c t a n (\frac{1}{n})$ $- a r c t a n (n) + a r c t a n (\frac{1}{n}) + 2 l n (n) - \frac{1}{2} (2 l n (n))$ $A_{n} = - \frac{a r c t a n (n)}{n} + n a r c t a n (\frac{1}{n}) + \frac{π}{2} - 2 a r c t a n (n)$ $+ l n (n)$ $2) w e h a v e$ ${l i m}_{n \to + \infty} (- \frac{a r c t a n (n)}{n} + n a r c t a n (\frac{1}{n}) + \frac{π}{2} - 2 a r c t s n (n))$ $= 0 + 1 - \frac{π}{2} - π b u t {l i m}_{n \to + \infty} l n (n) = + \infty \Rightarrow$ ${l i m}_{n \to + \infty} A_{n} = + \infty .$
	Answered by MJS last updated on 15/May/18

	$\int (1 + \frac{1}{x^{2}}) \arctan (x) d x =$ $= \int \arctan (x) d x + \int \frac{\arctan (x)}{x^{2}} d x =$ $\int \arctan (x) d x =$ $[\begin{matrix} \int u^{'} v = u v - \int {u v}^{'} \\ u^{'} = 1 \Rightarrow u = x \\ v = \arctan (x) \Rightarrow v^{'} = \frac{1}{x^{2} + 1} \end{matrix}]$ $= x \arctan (x) - \int \frac{x}{x^{2} + 1} d x =$ $[\begin{matrix} t = x^{2} + 1 \to d x = \frac{d t}{2 x} \end{matrix}]$ $= x \arctan (x) - \frac{1}{2} \int \frac{1}{t} d t =$ $= x \arctan (x) - \frac{\ln (t)}{2} =$ $= x \arctan (x) - \frac{\ln (x^{2} + 1)}{2}$ $\int \frac{\arctan (x)}{x^{2}} d x =$ $[\begin{matrix} \int u^{'} v = u v - \int {u v}^{'} \\ u^{'} = \frac{1}{x^{2}} \Rightarrow u = - \frac{1}{x} \\ v = \arctan (x) \Rightarrow v^{'} = \frac{1}{x^{2} + 1} \end{matrix}]$ $= - \frac{\arctan (x)}{x} + \int \frac{1}{x (x^{2} + 1)} d x =$ $= - \frac{\arctan (x)}{x} + \int (\frac{1}{x} - \frac{x^{2}}{x^{2} + 1}) d x =$ $[\begin{matrix} same as above \end{matrix}]$ $= - \frac{\arctan (x)}{x} + \ln (x) - \frac{\ln (x^{2} + 1)}{2}$ $= x \arctan (x) - \frac{\ln (x^{2} + 1)}{2} - \frac{\arctan (x)}{x} + \ln (x) - \frac{\ln (x^{2} + 1)}{2} =$ $= (\frac{x^{2} - 1}{x}) \arctan (x) + \ln (\frac{x}{x^{2} + 1}) + C$ $\underset{\frac{1}{n}}{\int^{n}} (1 + \frac{1}{x^{2}}) \arctan (x) d x =$ $= (\frac{n^{2} - 1}{n}) \arctan (n) + \ln (\frac{n}{n^{2} + 1}) - ((\frac{\frac{1}{n^{2}} - 1}{\frac{1}{n}}) \arctan (\frac{1}{n}) + \ln (\frac{\frac{1}{n}}{\frac{1}{n^{2}} + 1})) =$ $= (\frac{n^{2} - 1}{n}) \arctan (n) + \ln (\frac{n}{n^{2} + 1}) - ((\frac{1 - n^{2}}{n}) (\frac{π}{2} - \arctan (n)) + \ln (\frac{n}{n^{2} + 1})) =$ $= \frac{π (n^{2} - 1)}{2 n}$ $\lim_{n \to \infty} \frac{π (n^{2} - 1)}{2 n} = \frac{π}{2} \lim_{n \to \infty} (n - \frac{1}{n}) = \infty$
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