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let-F-x-x-1-x-arctan-t-t-dt-1-calculate-dF-dx-x-2-find-F-x-




Question Number 36438 by prof Abdo imad last updated on 02/Jun/18
let F(x) =∫_x ^(1/x)   ((arctan(t))/t)dt  1) calculate  (dF/dx)(x)  2) find F(x).
letF(x)=x1xarctan(t)tdt1)calculatedFdx(x)2)findF(x).
Commented by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18
Commented by prof Abdo imad last updated on 03/Jun/18
changement t =(1/u) give  2)F(x) = ∫_(1/x) ^x    (((π/2) −arctan(u))/(1/u)) (−(du/u^2 ))  = ∫_x ^(1/x)   (((π/2) −arctan(u))/u)du   = (π/2) ∫_x ^(1/x)   (du/u)  − ∫_x ^(1/x)    ((arctan(u))/u) du ⇒  2F(x) = (π/2)[ln∣u∣]_x ^(1/x)  =(π/2){ −2ln(x)}  = −π ln(x) ⇒ F(x) =−(π/2)ln(x)  1)  (dF/dx)(x) = −(π/(2x))  with x>0
changementt=1ugive2)F(x)=1xxπ2arctan(u)1u(duu2)=x1xπ2arctan(u)udu=π2x1xduux1xarctan(u)udu2F(x)=π2[lnu]x1x=π2{2ln(x)}=πln(x)F(x)=π2ln(x)1)dFdx(x)=π2xwithx>0
Commented by prof Abdo imad last updated on 03/Jun/18
add x>0 to Q.
addx>0toQ.
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18
F′(x)=∫_x ^(1/x) (∂/∂x)(((tan^(−1) t)/t))dt+((tan^(−1) ((1/x)))/(1/x))((d((1/x)))/dx)−  ((tan^(−1) (x))/x)(dx/dx)  =0+((tan^(−1) ((1/x)))/(1/x))×((−1)/x^2 )−((tan^(−1) (x))/x)  =((tan^(−1) ((1/x)))/(−x))−((tan^(−1) (x))/x)  =((−1)/x){((tan^(−1) ((1/x))+tan^(−1) (x))/1)}  =((−1)/x)tan^(−1) {((x+(1/x)))/(1−x.(1/x)))}  =(1/x)×((Π/2)]  (dF/dx)=(Π/2)(1/x)  dF=(Π/2)(dx/x)  F=(Π/2)lnx
F(x)=x1xx(tan1tt)dt+tan1(1x)1xd(1x)dxtan1(x)xdxdx=0+tan1(1x)1x×1x2tan1(x)x=tan1(1x)xtan1(x)x=1x{tan1(1x)+tan1(x)1}=1xtan1{x+1x)1x.1x}=1x×(Π2]dFdx=Π21xdF=Π2dxxF=Π2lnx

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