Question Number 49954 by maxmathsup by imad last updated on 12/Dec/18
$${find}\:\:\:{f}\left(\alpha\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left(\alpha{x}\right)}{\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left(\mathrm{2}{x}\right)}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\:\:{dx}\:\:{and}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left(\mathrm{3}{x}\right)}{\mathrm{1}+\mathrm{9}{x}^{\mathrm{2}} }\:{dx}\:. \\ $$
Commented by Abdo msup. last updated on 14/Dec/18
![1) by parts u^′ =(1/(1+α^2 x^2 )) and v=(1/α)arctan(αx) ⇒ f(α)=[(1/α) arctan(αx)arctan(αx)]_0 ^1 −∫_0 ^1 (1/α) arctan(αx)(α/(1+α^2 x^2 ))dx =(1/α) arctan^2 (α)−f(α) ⇒2f(α)=(1/α) arctan^2 (α) ⇒ f(α)=(1/(2α)) arctan^2 (α) 2)∫_0 ^1 ((arctan(2x))/(1+4x^2 ))dx =f(2)=(1/4) arctan^2 (2) ∫_0 ^1 ((arctan(3x))/(1+9x^2 ))dx =f(3)=(1/6)arctan^2 (3).](https://www.tinkutara.com/question/Q50129.png)
$$\left.\mathrm{1}\right)\:{by}\:{parts}\:{u}^{'} =\frac{\mathrm{1}}{\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} }\:{and}\:{v}=\frac{\mathrm{1}}{\alpha}{arctan}\left(\alpha{x}\right)\:\Rightarrow \\ $$$${f}\left(\alpha\right)=\left[\frac{\mathrm{1}}{\alpha}\:{arctan}\left(\alpha{x}\right){arctan}\left(\alpha{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{\alpha}\:{arctan}\left(\alpha{x}\right)\frac{\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\alpha}\:{arctan}^{\mathrm{2}} \left(\alpha\right)−{f}\left(\alpha\right)\:\Rightarrow\mathrm{2}{f}\left(\alpha\right)=\frac{\mathrm{1}}{\alpha}\:{arctan}^{\mathrm{2}} \left(\alpha\right)\:\Rightarrow \\ $$$${f}\left(\alpha\right)=\frac{\mathrm{1}}{\mathrm{2}\alpha}\:{arctan}^{\mathrm{2}} \left(\alpha\right) \\ $$$$\left.\mathrm{2}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left(\mathrm{2}{x}\right)}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }{dx}\:={f}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{4}}\:{arctan}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left(\mathrm{3}{x}\right)}{\mathrm{1}+\mathrm{9}{x}^{\mathrm{2}} }{dx}\:={f}\left(\mathrm{3}\right)=\frac{\mathrm{1}}{\mathrm{6}}{arctan}^{\mathrm{2}} \left(\mathrm{3}\right). \\ $$
Answered by Smail last updated on 13/Dec/18
![f(α)=∫_0 ^1 ((arctan(αx))/(1+α^2 x^2 ))dx =∫_0 ^1 ((d(arctan(αx))/dx)×arctan(αx)dx =(1/2)[(arctan(αx))^2 ]_0 ^1 (∫u′u^n dx=(1/(n+1))u^(n+1) +c) f(α)=(1/2)arctan^2 (α) f(2)=(1/2)arctan^2 (2)](https://www.tinkutara.com/question/Q50063.png)
$${f}\left(\alpha\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{arctan}\left(\alpha{x}\right)}{\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{d}\left({arctan}\left(\alpha{x}\right)\right.}{{dx}}×{arctan}\left(\alpha{x}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\left({arctan}\left(\alpha{x}\right)\right)^{\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:\:\left(\int{u}'{u}^{{n}} {dx}=\frac{\mathrm{1}}{{n}+\mathrm{1}}{u}^{{n}+\mathrm{1}} +{c}\right) \\ $$$${f}\left(\alpha\right)=\frac{\mathrm{1}}{\mathrm{2}}{arctan}^{\mathrm{2}} \left(\alpha\right) \\ $$$${f}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{2}}{arctan}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$