Question Number 46855 by maxmathsup by imad last updated on 01/Nov/18
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}\:{arctan}\left(\mathrm{1}+{x}\right){dx} \\ $$
Commented by maxmathsup by imad last updated on 02/Nov/18
![let I = ∫_0 ^1 x arctan(1+x)dx by parts u^′ =x and v =arctan(1+x) I =[(x^2 /2) arctan(1+x)]_0 ^1 −∫_0 ^1 (x^2 /2) (1/(1+(1+x)^2 ))dx =(1/2) arctan(2)−(1/2) ∫_0 ^1 ((x^2 dx)/(1+(1+x)^2 )) but ∫_0 ^1 (x^2 /(1+(1+x)^2 ))dx =_(1+x =u) ∫_1 ^2 (((u−1)^2 )/(1+u^2 )) dx =∫_1 ^2 ((u^2 −2u +1)/(1+u^2 ))du =∫_1 ^2 du −∫_1 ^2 ((2u)/(1+u^2 ))du =1 −[ln(1+u^2 )]_1 ^2 =1−(ln(5)−ln(2)) =1−ln((5/2)) ⇒ I =((arctan(2))/2) −(1/2) +(1/2)ln((5/2))](https://www.tinkutara.com/question/Q46901.png)
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}\:{arctan}\left(\mathrm{1}+{x}\right){dx}\:{by}\:{parts}\:\:{u}^{'} \:={x}\:{and}\:{v}\:={arctan}\left(\mathrm{1}+{x}\right) \\ $$$${I}\:=\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:{arctan}\left(\mathrm{1}+{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{x}^{\mathrm{2}} {dx}}{\mathrm{1}+\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\:{but}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}+\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx}\:=_{\mathrm{1}+{x}\:={u}} \:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{\left({u}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\:{dx}\:=\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{{u}^{\mathrm{2}} −\mathrm{2}{u}\:+\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} \:{du}\:−\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }{du}\:=\mathrm{1}\:−\left[{ln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\right]_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{1}−\left({ln}\left(\mathrm{5}\right)−{ln}\left(\mathrm{2}\right)\right) \\ $$$$=\mathrm{1}−{ln}\left(\frac{\mathrm{5}}{\mathrm{2}}\right)\:\Rightarrow\:{I}\:=\frac{{arctan}\left(\mathrm{2}\right)}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{5}}{\mathrm{2}}\right) \\ $$