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1-calculate-F-x-1-x-arctan-t-t-2-dt-with-x-1-2-calculate-A-n-1-n-arctan-t-t-2-dt-and-find-lim-n-A-n-




Question Number 39369 by maxmathsup by imad last updated on 05/Jul/18
1) calculate F(x)= ∫_1 ^(√x)    ((arctan(t))/t^2 )dt   with x≥1  2) calculate   A_n = ∫_1 ^(√n)   ((arctan(t))/t^2 ) dt  and find lim_(n→+∞)  A_n
1)calculateF(x)=1xarctan(t)t2dtwithx12)calculateAn=1narctan(t)t2dtandfindlimn+An
Commented by math khazana by abdo last updated on 07/Jul/18
1) by parts F(x) =[−(1/t) arctant]_1 ^(√x)  −∫_1 ^(√x) ((−1)/(t(1+t^2 )))dt  =(π/4) −((arctan((√x)))/( (√x))) + ∫_1 ^(√x) ( (1/t) −(t/(1+t^2 )))dt  =(π/4) −((arctan((√x)))/( (√x)))  +ln((√x)) −(1/2)[ln(1+t^2 )]_1 ^(√x)   =(π/4) −((arctan((√x)))/( (√x))) +ln((√x)) −(1/2){ln(1+x)−ln(2)}  2) we have A_n =F(n)  =(π/4) −((arctan((√n)))/( (√n))) + ln((√n)) −ln(√(1+n)) +((ln(2))/2)  =(π/4) +((ln(2))/2)  +ln(((√n)/( (√(n+1))))) −((arctan((√(n))))/( (√n))) ⇒  lim_(n→+∞)  A_n =(π/4) +((ln(2))/2) .
1)bypartsF(x)=[1tarctant]1x1x1t(1+t2)dt=π4arctan(x)x+1x(1tt1+t2)dt=π4arctan(x)x+ln(x)12[ln(1+t2)]1x=π4arctan(x)x+ln(x)12{ln(1+x)ln(2)}2)wehaveAn=F(n)=π4arctan(n)n+ln(n)ln1+n+ln(2)2=π4+ln(2)2+ln(nn+1)arctan(n)nlimn+An=π4+ln(2)2.

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