Question Number 36438 by prof Abdo imad last updated on 02/Jun/18

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](https://www.tinkutara.com/question/Q36520.png)
Commented by prof Abdo imad last updated on 03/Jun/18

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](https://www.tinkutara.com/question/Q36471.png)