Question Number 35685 by prof Abdo imad last updated on 22/May/18
$${calculate}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:{x}\:{artan}\left(\mathrm{2}{x}+\mathrm{1}\right){dx} \\ $$
Commented by prof Abdo imad last updated on 23/May/18
![let put I = ∫_0 ^(π/4) x arctan(2x+1)dx by parts I = [ (x^2 /2)arctan(2x+1)]_0 ^(π/4) − ∫_0 ^(π/4) (x^2 /2) (2/(1 +(2x+1)^2 ))dx =(π^2 /(32))arctan((π/2)+1) −∫_0 ^(π/4) (x^2 /(1+(2x+1)^2 ))dx let calculate J= ∫_0 ^(π/4) (x^2 /(1+(2x+1)^2 ))dx J =_(2x+1=u) ∫_1 ^((π/2)+1) (((((u−1)/2))^2 )/(1+u^2 )) (du/2) = (1/8) ∫_1 ^((π/2)+1) ((u^2 −2u +1)/(1+u^2 ))du 8J = ∫_1 ^((π/2)+1) (1 −((2u)/(1+u^2 )))du = (π/2) −[ ln(1+u^2 )]_1 ^((π/2)+1) =(π/2) −{ ln(1+((π/2)+1)^2 ) −ln(2)} ⇒ J = (π/(16)) −(1/8){ln(1+((π/2)+1)^2 ) −ln(2)} so I = (π^2 /(32)) arctan((π/2)+1) −(π/(16)) +(1/8){ln(1+((π/2)+1)^2 )−ln(2)}](https://www.tinkutara.com/question/Q35795.png)
$${let}\:{put}\:{I}\:\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{x}\:{arctan}\left(\mathrm{2}{x}+\mathrm{1}\right){dx}\:{by}\:{parts} \\ $$$${I}\:=\:\left[\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{arctan}\left(\mathrm{2}{x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:−\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\frac{\mathrm{2}}{\mathrm{1}\:+\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{32}}{arctan}\left(\frac{\pi}{\mathrm{2}}+\mathrm{1}\right)\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}+\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:{let}\: \\ $$$${calculate}\:{J}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}+\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$${J}\:=_{\mathrm{2}{x}+\mathrm{1}={u}} \:\:\:\int_{\mathrm{1}} ^{\frac{\pi}{\mathrm{2}}+\mathrm{1}} \:\:\:\frac{\left(\frac{{u}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{{du}}{\mathrm{2}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{8}}\:\int_{\mathrm{1}} ^{\frac{\pi}{\mathrm{2}}+\mathrm{1}} \:\:\:\:\:\frac{{u}^{\mathrm{2}} \:−\mathrm{2}{u}\:+\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$$\mathrm{8}{J}\:=\:\int_{\mathrm{1}} ^{\frac{\pi}{\mathrm{2}}+\mathrm{1}} \:\:\:\left(\mathrm{1}\:\:−\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\right){du} \\ $$$$=\:\frac{\pi}{\mathrm{2}}\:−\left[\:{ln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\right]_{\mathrm{1}} ^{\frac{\pi}{\mathrm{2}}+\mathrm{1}} \\ $$$$=\frac{\pi}{\mathrm{2}}\:−\left\{\:{ln}\left(\mathrm{1}+\left(\frac{\pi}{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} \right)\:−{ln}\left(\mathrm{2}\right)\right\}\:\Rightarrow \\ $$$${J}\:=\:\frac{\pi}{\mathrm{16}}\:−\frac{\mathrm{1}}{\mathrm{8}}\left\{{ln}\left(\mathrm{1}+\left(\frac{\pi}{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} \right)\:−{ln}\left(\mathrm{2}\right)\right\}\:{so} \\ $$$${I}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{32}}\:{arctan}\left(\frac{\pi}{\mathrm{2}}+\mathrm{1}\right)\:−\frac{\pi}{\mathrm{16}}\:+\frac{\mathrm{1}}{\mathrm{8}}\left\{{ln}\left(\mathrm{1}+\left(\frac{\pi}{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} \right)−{ln}\left(\mathrm{2}\right)\right\} \\ $$
Answered by ajfour last updated on 23/May/18
![I=(x^2 /2)tan^(−1) (2x+1)∣_0 ^(π/4) −∫_0 ^( π/4) ((x^2 /2))[(2/(1+(2x+1)^2 ))]dx =(π^2 /(32))tan^(−1) (1+(π/2))−I_1 I_1 =(1/4)∫_0 ^( π/4) ((1+(2x+1)^2 −2(2x+1))/(1+(2x+1)^2 ))dx =(x/4)∣_0 ^(π/4) −(1/4)∫_( 2) ^( 1+(1+(π/2))^2 ) (dt/t) =(π/(16))−(1/4)ln [(1/2)+(((π+2)^2 )/8)] I = (π^2 /(32))tan^(−1) (1+(π/2))−(π/(16)) +(1/4)ln [(1/2)+(((π+2)^2 )/8)] .](https://www.tinkutara.com/question/Q35803.png)
$${I}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{x}+\mathrm{1}\right)\mid_{\mathrm{0}} ^{\pi/\mathrm{4}} −\int_{\mathrm{0}} ^{\:\:\pi/\mathrm{4}} \:\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\left[\frac{\mathrm{2}}{\mathrm{1}+\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }\right]{dx} \\ $$$$\:=\frac{\pi^{\mathrm{2}} }{\mathrm{32}}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}+\frac{\pi}{\mathrm{2}}\right)−{I}_{\mathrm{1}} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\:\pi/\mathrm{4}} \frac{\mathrm{1}+\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}{x}+\mathrm{1}\right)}{\mathrm{1}+\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$\:\:\:\:=\frac{{x}}{\mathrm{4}}\mid_{\mathrm{0}} ^{\pi/\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{4}}\int_{\:\mathrm{2}} ^{\:\:\mathrm{1}+\left(\mathrm{1}+\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} } \:\frac{{dt}}{{t}} \\ $$$$\:\:\:=\frac{\pi}{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\left[\frac{\mathrm{1}}{\mathrm{2}}+\frac{\left(\pi+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{8}}\right] \\ $$$$\:\:\:\:\:\:{I}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{32}}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}+\frac{\pi}{\mathrm{2}}\right)−\frac{\pi}{\mathrm{16}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\left[\frac{\mathrm{1}}{\mathrm{2}}+\frac{\left(\pi+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{8}}\right]\:. \\ $$