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let-A-n-1-n-n-1-1-x-2-arctanx-dx-1-calculate-A-n-2-find-lim-n-A-n-

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Question Number 35049 by math khazana by abdo last updated on 14/May/18
let A_n = ∫_(1/n) ^n (1+(1/x^2 ))arctanx dx 1) calculate A_n 2) find lim_(n→+∞) A_n
letAn=∫1nn(1+1x2)arctanxdx1)calculateAn2)findlimn→+∞An
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)letintegratebypartsu′=1+1x2andv=arctanxAn=[(1−1x)arctanx]1nn−∫1nn(1−1x)dx1+x2=(1−1n)arctan(n)−(1−n)arctan(1n)−∫1nndx1+x2+∫1nndxx(1+x2)but∫1nndx1+x2=arctan(n)−arctan(1n)∫1nn1x(1+x2)dx=∫1nn(1x−x1+x2)dx=ln(n)−ln(1n)−12[ln(1+x2)]1nn=2ln(n)−12{ln(1+n2)−ln(1+1n2)}⇒An=(1−1n)arctan(n)−(1−n)arctan(1n)−arctan(n)+arctan(1n)+2ln(n)−12(2ln(n))An=−arctan(n)n+narctan(1n)+π2−2arctan(n)+ln(n)2)wehavelimn→+∞(−arctan(n)n+narctan(1n)+π2−2arctsn(n))=0+1−π2−πbutlimn→+∞ln(n)=+∞⇒limn→+∞An=+∞.
Answered by MJS last updated on 15/May/18
∫(1+(1/x^2 ))arctan(x)dx= =∫arctan(x)dx+∫((arctan(x))/x^2 )dx= ∫arctan(x)dx= [((∫u′v=uv−∫uv′)),((u′=1 ⇒ u=x)),((v=arctan(x) ⇒ v′=(1/(x^2 +1)))) ] =xarctan(x)−∫(x/(x^2 +1))dx= [((t=x^2 +1 → dx=(dt/(2x)))) ] =xarctan(x)−(1/2)∫(1/t)dt= =xarctan(x)−((ln(t))/2)= =xarctan(x)−((ln(x^2 +1))/2) ∫((arctan(x))/x^2 )dx= [((∫u′v=uv−∫uv′)),((u′=(1/x^2 ) ⇒ u=−(1/x))),((v=arctan(x) ⇒ v′=(1/(x^2 +1)))) ] =−((arctan(x))/x)+∫(1/(x(x^2 +1)))dx= =−((arctan(x))/x)+∫((1/x)−(x^2 /(x^2 +1)))dx= [((same as above)) ] =−((arctan(x))/x)+ln(x)−((ln(x^2 +1))/2) =xarctan(x)−((ln(x^2 +1))/2)−((arctan(x))/x)+ln(x)−((ln(x^2 +1))/2)= =(((x^2 −1)/x))arctan(x)+ln((x/(x^2 +1)))+C ∫_(1/n) ^n (1+(1/x^2 ))arctan(x)dx= =(((n^2 −1)/n))arctan(n)+ln((n/(n^2 +1)))−(((((1/n^2 )−1)/(1/n)))arctan((1/n))+ln(((1/n)/((1/n^2 )+1))))= =(((n^2 −1)/n))arctan(n)+ln((n/(n^2 +1)))−((((1−n^2 )/n))((π/2)−arctan(n))+ln((n/(n^2 +1))))= =((π(n^2 −1))/(2n)) lim_(n→∞) ((π(n^2 −1))/(2n))=(π/2)lim_(n→∞) (n−(1/n))=∞
∫(1+1x2)arctan(x)dx==∫arctan(x)dx+∫arctan(x)x2dx=∫arctan(x)dx=[∫u′v=uv−∫uv′u′=1⇒u=xv=arctan(x)⇒v′=1x2+1]=xarctan(x)−∫xx2+1dx=[t=x2+1→dx=dt2x]=xarctan(x)−12∫1tdt==xarctan(x)−ln(t)2==xarctan(x)−ln(x2+1)2∫arctan(x)x2dx=[∫u′v=uv−∫uv′u′=1x2⇒u=−1xv=arctan(x)⇒v′=1x2+1]=−arctan(x)x+∫1x(x2+1)dx==−arctan(x)x+∫(1x−x2x2+1)dx=[sameasabove]=−arctan(x)x+ln(x)−ln(x2+1)2=xarctan(x)−ln(x2+1)2−arctan(x)x+ln(x)−ln(x2+1)2==(x2−1x)arctan(x)+ln(xx2+1)+C∫n1n(1+1x2)arctan(x)dx==(n2−1n)arctan(n)+ln(nn2+1)−((1n2−11n)arctan(1n)+ln(1n1n2+1))==(n2−1n)arctan(n)+ln(nn2+1)−((1−n2n)(π2−arctan(n))+ln(nn2+1))==π(n2−1)2nlimn→∞π(n2−1)2n=π2limn→∞(n−1n)=∞

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