Question Number 62908 by aliesam last updated on 26/Jun/19
$$\int\frac{{arctan}\left({x}\right)}{{x}}{dx} \\ $$
Commented by mathmax by abdo last updated on 26/Jun/19
![for all x from R u→((arctanu)/u)is integrable on ]0,x]let f(t) =∫_0 ^x ((arctan(tu))/u)du we have f^′ (t) =∫_0 ^x arctan(tu)du =_(by parts) [u arctan(tu)]_(u=0) ^(u=x) −∫_0 ^x u(t/(1+t^2 u^2 )) du =x arctan(tx)−(1/(2t)) ∫_0 ^x ((2t^2 u)/(1+t^2 u^2 )) du =xarctan(tx)−(1/(2t))[ ln(1+t^2 u^2 )]_(u=0) ^(u=x) =xarctan(tx)−(1/(2t))ln(1+t^2 x^2 ) ⇒ f(t) =∫_0 ^t x arctan(ux)du −∫_0 ^t ((ln(1+x^2 u^2 ))/(2u)) du +C ⇒ ∫_0 ^x ((arctan(u))/u) du =f(1) =x∫_0 ^1 arctan(ux)du−∫_0 ^1 ((ln(1+x^2 u^2 ))/(2u)) du +C x=0 ⇒C =0 ⇒∫_0 ^x ((arctan(u))/u) du=x ∫_0 ^1 arctan(ux)du−∫_0 ^1 ((ln(1+x^2 u^2 3)/(2u)) du ...be continued....](https://www.tinkutara.com/question/Q62913.png)
$$\left.\:\left.{for}\:{all}\:{x}\:{from}\:{R}\:\:\:\:{u}\rightarrow\frac{{arctanu}}{{u}}{is}\:{integrable}\:{on}\:\right]\mathrm{0},{x}\right]{let}\:{f}\left({t}\right)\:=\int_{\mathrm{0}} ^{{x}} \:\frac{{arctan}\left({tu}\right)}{{u}}{du} \\ $$$${we}\:{have}\:{f}^{'} \left({t}\right)\:=\int_{\mathrm{0}} ^{{x}} \:{arctan}\left({tu}\right){du}\:=_{{by}\:{parts}} \:\:\:\left[{u}\:{arctan}\left({tu}\right)\right]_{{u}=\mathrm{0}} ^{{u}={x}} \:−\int_{\mathrm{0}} ^{{x}} \:{u}\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} {u}^{\mathrm{2}} }\:{du} \\ $$$$={x}\:{arctan}\left({tx}\right)−\frac{\mathrm{1}}{\mathrm{2}{t}}\:\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{\mathrm{2}{t}^{\mathrm{2}} {u}}{\mathrm{1}+{t}^{\mathrm{2}} {u}^{\mathrm{2}} }\:{du} \\ $$$$={xarctan}\left({tx}\right)−\frac{\mathrm{1}}{\mathrm{2}{t}}\left[\:{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} {u}^{\mathrm{2}} \right)\right]_{{u}=\mathrm{0}} ^{{u}={x}} \:={xarctan}\left({tx}\right)−\frac{\mathrm{1}}{\mathrm{2}{t}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${f}\left({t}\right)\:=\int_{\mathrm{0}} ^{{t}} \:\:{x}\:{arctan}\left({ux}\right){du}\:−\int_{\mathrm{0}} ^{{t}} \:\frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} \right)}{\mathrm{2}{u}}\:{du}\:+{C}\:\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{{x}} \:\frac{{arctan}\left({u}\right)}{{u}}\:{du}\:={f}\left(\mathrm{1}\right)\:={x}\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left({ux}\right){du}−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} \right)}{\mathrm{2}{u}}\:{du}\:+{C} \\ $$$${x}=\mathrm{0}\:\Rightarrow{C}\:=\mathrm{0}\:\Rightarrow\int_{\mathrm{0}} ^{{x}} \:\frac{{arctan}\left({u}\right)}{{u}}\:{du}={x}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left({ux}\right){du}−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} \mathrm{3}\right.}{\mathrm{2}{u}}\:{du} \\ $$$$…{be}\:{continued}…. \\ $$