Question Number 36438 by prof Abdo imad last updated on 02/Jun/18
$${let}\:{F}\left({x}\right)\:=\int_{{x}} ^{\frac{\mathrm{1}}{{x}}} \:\:\frac{{arctan}\left({t}\right)}{{t}}{dt} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\:\frac{{dF}}{{dx}}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{F}\left({x}\right). \\ $$
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)
$${changement}\:{t}\:=\frac{\mathrm{1}}{{u}}\:{give} \\ $$$$\left.\mathrm{2}\right){F}\left({x}\right)\:=\:\int_{\frac{\mathrm{1}}{{x}}} ^{{x}} \:\:\:\frac{\frac{\pi}{\mathrm{2}}\:−{arctan}\left({u}\right)}{\frac{\mathrm{1}}{{u}}}\:\left(−\frac{{du}}{{u}^{\mathrm{2}} }\right) \\ $$$$=\:\int_{{x}} ^{\frac{\mathrm{1}}{{x}}} \:\:\frac{\frac{\pi}{\mathrm{2}}\:−{arctan}\left({u}\right)}{{u}}{du}\: \\ $$$$=\:\frac{\pi}{\mathrm{2}}\:\int_{{x}} ^{\frac{\mathrm{1}}{{x}}} \:\:\frac{{du}}{{u}}\:\:−\:\int_{{x}} ^{\frac{\mathrm{1}}{{x}}} \:\:\:\frac{{arctan}\left({u}\right)}{{u}}\:{du}\:\Rightarrow \\ $$$$\mathrm{2}{F}\left({x}\right)\:=\:\frac{\pi}{\mathrm{2}}\left[{ln}\mid{u}\mid\right]_{{x}} ^{\frac{\mathrm{1}}{{x}}} \:=\frac{\pi}{\mathrm{2}}\left\{\:−\mathrm{2}{ln}\left({x}\right)\right\} \\ $$$$=\:−\pi\:{ln}\left({x}\right)\:\Rightarrow\:{F}\left({x}\right)\:=−\frac{\pi}{\mathrm{2}}{ln}\left({x}\right) \\ $$$$\left.\mathrm{1}\right)\:\:\frac{{dF}}{{dx}}\left({x}\right)\:=\:−\frac{\pi}{\mathrm{2}{x}}\:\:{with}\:{x}>\mathrm{0} \\ $$
Commented by prof Abdo imad last updated on 03/Jun/18
$${add}\:{x}>\mathrm{0}\:{to}\:{Q}. \\ $$
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)
$${F}'\left({x}\right)=\int_{{x}} ^{\frac{\mathrm{1}}{{x}}} \frac{\partial}{\partial{x}}\left(\frac{{tan}^{−\mathrm{1}} {t}}{{t}}\right){dt}+\frac{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)}{\frac{\mathrm{1}}{{x}}}\frac{{d}\left(\frac{\mathrm{1}}{{x}}\right)}{{dx}}− \\ $$$$\frac{{tan}^{−\mathrm{1}} \left({x}\right)}{{x}}\frac{{dx}}{{dx}} \\ $$$$=\mathrm{0}+\frac{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)}{\frac{\mathrm{1}}{{x}}}×\frac{−\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{{tan}^{−\mathrm{1}} \left({x}\right)}{{x}} \\ $$$$=\frac{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)}{−{x}}−\frac{{tan}^{−\mathrm{1}} \left({x}\right)}{{x}} \\ $$$$=\frac{−\mathrm{1}}{{x}}\left\{\frac{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)+{tan}^{−\mathrm{1}} \left({x}\right)}{\mathrm{1}}\right\} \\ $$$$=\frac{−\mathrm{1}}{{x}}{tan}^{−\mathrm{1}} \left\{\frac{\left.{x}+\frac{\mathrm{1}}{{x}}\right)}{\mathrm{1}−{x}.\frac{\mathrm{1}}{{x}}}\right\} \\ $$$$=\frac{\mathrm{1}}{{x}}×\left(\frac{\Pi}{\mathrm{2}}\right] \\ $$$$\frac{{dF}}{{dx}}=\frac{\Pi}{\mathrm{2}}\frac{\mathrm{1}}{{x}} \\ $$$${dF}=\frac{\Pi}{\mathrm{2}}\frac{{dx}}{{x}} \\ $$$${F}=\frac{\Pi}{\mathrm{2}}{lnx} \\ $$