find U_n =∫_1 ^n arctan(x+(1/x))dx and determine nature of Σ U


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Question Number 81434 by abdomathmax last updated on 13/Feb/20

$f i n d U_{n} = \int_{1}^{n} a r c t a n (x + \frac{1}{x}) d x$ $a n d d e t e r m i n e n a t u r e o f Σ U_{n}$
Commented by abdomathmax last updated on 13/Feb/20
$by parts U_n =[xarctan(x+(1/x))]_1 ^n −∫_1 ^n x×((1−(1/x^2 ))/(1+(x+(1/x))^2 ))dx =n arctan(n+(1/n))−arctan(2)−∫_1 ^n x×((x^2 −1)/(x^2 {1+(((x^2 +1)^2 )/x^2 )}))dx =narctan(n+(1/n))−arctan(2)−∫_1 ^n ((x^3 −x)/(x^2 +(x^2 +1)^2 ))dx =narctan(n+(1/n))−arctan(2)−∫_1 ^n ((x^3 −x)/(x^2 +x^4 +2x^2 +1))dx =narctan(n+(1/n))−arctan(2)−∫_1 ^n ((x^3 −x)/(x^4 +3x^2 +1))dx x^4 +3x^2 +1=0 →u^2 +3u +1=0(u=x^2 ) ⇒ Δ=9−4=5 ⇒u_1 =((−3+(√5))/2) and u_2 =((−3−(√5))/2) ⇒F(x)=((x^3 −x)/(x^4 +3x^2 +1)) =((x^3 −x)/((x^2 −u_1 )(x^2 −u_2 ))) =((x^3 −x)/(u_1 −u_2 ))((1/(x^2 −u_1 ))−(1/(x^2 −u_2 ))) ⇒ ∫_1 ^n F(x)dx =(1/(√5))∫_1 ^n ((x^3 −x)/(x^2 −u_1 ))dx−(1/(√5)) ∫_1 ^n ((x^3 −x)/(x^2 −u_2 ))dx wehavd ∫_1 ^n ((x^3 −x)/(x^2 −u_1 ))dx =∫_1 ^n ((x(x^2 −u_1 )+xu_1 −x)/(x^2 −u_1 ))dx =(x^2 /2)]_1 ^n +∫_1 ^n (((u_1 −1)x)/(x^2 −u_1 ))dx=(x^2 /2) +((u_1 −1)/2)ln∣x^2 −u_1 ∣ =(n^2 /2)−(1/2) +((u_1 −1)/2)[ln∣x^2 −u_1 ∣]_1 ^n =((n^2 −1)/2) +((u_1 −1)/2){ln(n^2 −u_1 )−ln∣1−u_1 ∣}also we have ∫_1 ^n =((n^2 −1)/2)+((u_2 −1)/2){ln(n^2 −u_2 )−ln∣1−u_2 )} ⇒ U_n =narctan(1+(1/n))−arctan2 −(1/(√5)){((u_1 −1)/2)ln(n^2 −u_1 )−((u_2 −1)/2)ln(n^2 −u_2 ) +((u_2 −1)/2)ln∣1−u_2 ∣−((u_1 −1)/2)ln∣1−u_1 ∣}$
$b y p a r t s U_{n} = {[x a r c t a n (x + \frac{1}{x})]}_{1}^{n} - \int_{1}^{n} x \times \frac{1 - \frac{1}{x^{2}}}{1 + {(x + \frac{1}{x})}^{2}} d x$ $= n a r c t a n (n + \frac{1}{n}) - a r c t a n (2) - \int_{1}^{n} x \times \frac{x^{2} - 1}{x^{2} {1 + \frac{{(x^{2} + 1)}^{2}}{x^{2}}}} d x$ $= n a r c t a n (n + \frac{1}{n}) - a r c t a n (2) - \int_{1}^{n} \frac{x^{3} - x}{x^{2} + {(x^{2} + 1)}^{2}} d x$ $= n a r c t a n (n + \frac{1}{n}) - a r c t a n (2) - \int_{1}^{n} \frac{x^{3} - x}{x^{2} + x^{4} + 2 x^{2} + 1} d x$ $= n a r c t a n (n + \frac{1}{n}) - a r c t a n (2) - \int_{1}^{n} \frac{x^{3} - x}{x^{4} + 3 x^{2} + 1} d x$ $x^{4} + 3 x^{2} + 1 = 0 \to u^{2} + 3 u + 1 = 0 (u = x^{2}) \Rightarrow$ $Δ = 9 - 4 = 5 \Rightarrow u_{1} = \frac{- 3 + \sqrt{5}}{2} a n d u_{2} = \frac{- 3 - \sqrt{5}}{2}$ $\Rightarrow F (x) = \frac{x^{3} - x}{x^{4} + 3 x^{2} + 1} = \frac{x^{3} - x}{(x^{2} - u_{1}) (x^{2} - u_{2})}$ $= \frac{x^{3} - x}{u_{1} - u_{2}} (\frac{1}{x^{2} - u_{1}} - \frac{1}{x^{2} - u_{2}}) \Rightarrow$ $\int_{1}^{n} F (x) d x = \frac{1}{\sqrt{5}} \int_{1}^{n} \frac{x^{3} - x}{x^{2} - u_{1}} d x - \frac{1}{\sqrt{5}} \int_{1}^{n} \frac{x^{3} - x}{x^{2} - u_{2}} d x w e h a v d$ $\int_{1}^{n} \frac{x^{3} - x}{x^{2} - u_{1}} d x = \int_{1}^{n} \frac{x (x^{2} - u_{1}) + {x u}_{1} - x}{x^{2} - u_{1}} d x$ ${= \frac{x^{2}}{2}]}_{1}^{n} + \int_{1}^{n} \frac{(u_{1} - 1) x}{x^{2} - u_{1}} d x = \frac{x^{2}}{2} + \frac{u_{1} - 1}{2} l n ∣ x^{2} - u_{1} ∣$ $= \frac{n^{2}}{2} - \frac{1}{2} + \frac{u_{1} - 1}{2} {[l n ∣ x^{2} - u_{1} ∣]}_{1}^{n}$ $= \frac{n^{2} - 1}{2} + \frac{u_{1} - 1}{2} {l n (n^{2} - u_{1}) - l n ∣ 1 - u_{1} ∣} a l s o$ $w e h a v e$ $\int_{1}^{n} = \frac{n^{2} - 1}{2} + \frac{u_{2} - 1}{2} {l n (n^{2} - u_{2}) - l n ∣ 1 - u_{2})} \Rightarrow$ $U_{n} = n a r c t a n (1 + \frac{1}{n}) - a r c t a n 2$ $- \frac{1}{\sqrt{5}} {\frac{u_{1} - 1}{2} l n (n^{2} - u_{1}) - \frac{u_{2} - 1}{2} l n (n^{2} - u_{2})$ $+ \frac{u_{2} - 1}{2} l n ∣ 1 - u_{2} ∣ - \frac{u_{1} - 1}{2} l n ∣ 1 - u_{1} ∣}$
Commented by abdomathmax last updated on 13/Feb/20

$U_{n} \to + \infty a n d Σ U_{n} d i v e r g e s$
	Answered by mind is power last updated on 13/Feb/20

	$U_{n} ⩾ \int_{1}^{n} a r v t a n (x) d x$ $\Rightarrow U n ⩾ {[x a r c t a n (x) - \frac{l n (1 + x^{2})}{2}]}_{1}^{n}$ $U_{n} ⩾ n a r c t a n (n) - \frac{l n (1 + n^{2})}{2} - \frac{π}{4} + \frac{l n (2)}{2}$ $n a r c t a n (n) - \frac{l n (1 + n^{2})}{2} = n (a r c t a n (n) - \frac{l n (1 + n^{2})}{2 n}) + \frac{l n (2)}{2} - \frac{π}{4} \to \infty$ $\Rightarrow Σ U_{n} d i v e r g e$
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