Question Number 59274 by Mr X pcx last updated on 07/May/19
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right){dx} \\ $$
Commented by maxmathsup by imad last updated on 18/May/19
![we have I =∫_0 ^1 arctan(1+x+x^2 ) =∫_0 ^1 arcotan((1/(1+x+x^2 ))) =∫_0 ^1 arcotan(((x+1−x)/(1+(x+1)x)))dx let x+1=tanα and x =tanβ ⇒ I =∫_0 ^1 arcotan(((tanα−tanβ)/(1+tanα.tanβ)))dx =∫_0 ^1 arcotan(tan(α−β) =∫_0 ^1 arcotan(cotan((π/2)−α +β))dx =∫_0 ^1 ((π/2)−α +β)dx =(π/2) −∫_0 ^1 arctan(x+1) dx +∫_0 ^1 arctanx dx by parts ∫_0 ^1 arctanx dx =[x arctanx]_0 ^1 −∫_0 ^1 (x/(1+x^2 ))dx =(π/4) −[(1/2)ln(1+x^2 )]_0 ^1 =(π/4) −((ln(2))/2) ∫_0 ^1 arctan(x+1)dx =∫_1 ^2 arctan(x)dx =[xarctanx]_1 ^2 −∫_1 ^2 (x/(1+x^2 ))dx =2arctan2−(π/4) −(1/2)[ln(1+x^2 )]_1 ^2 =2arctan(2)−(π/4) −(1/2)(ln(5)−ln(2)) ⇒ I = (π/2) −2arctan(2)+(π/4) +(1/2)ln(5)−(1/2)ln(2) +((.π)/4) −((ln(2))/2) I =π −2arctan(2) +((ln(5)−ln(2))/2) .](https://www.tinkutara.com/question/Q60187.png)
$${we}\:{have}\:\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{arcotan}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}+{x}^{\mathrm{2}} }\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:{arcotan}\left(\frac{{x}+\mathrm{1}−{x}}{\mathrm{1}+\left({x}+\mathrm{1}\right){x}}\right){dx}\:{let}\:\:\:{x}+\mathrm{1}={tan}\alpha\:\:{and}\:{x}\:={tan}\beta\:\Rightarrow \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{arcotan}\left(\frac{{tan}\alpha−{tan}\beta}{\mathrm{1}+{tan}\alpha.{tan}\beta}\right){dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{arcotan}\left({tan}\left(\alpha−\beta\right)\right. \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:{arcotan}\left({cotan}\left(\frac{\pi}{\mathrm{2}}−\alpha\:+\beta\right)\right){dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\pi}{\mathrm{2}}−\alpha\:+\beta\right){dx} \\ $$$$=\frac{\pi}{\mathrm{2}}\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left({x}+\mathrm{1}\right)\:{dx}\:+\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctanx}\:{dx}\:\:\:{by}\:{parts} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctanx}\:{dx}\:=\left[{x}\:{arctanx}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\frac{\pi}{\mathrm{4}}\:−\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\pi}{\mathrm{4}}\:−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left({x}+\mathrm{1}\right){dx}\:=\int_{\mathrm{1}} ^{\mathrm{2}} \:{arctan}\left({x}\right){dx}\:=\left[{xarctanx}\right]_{\mathrm{1}} ^{\mathrm{2}} \:−\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\mathrm{2}{arctan}\mathrm{2}−\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right]_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{2}{arctan}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\mathrm{5}\right)−{ln}\left(\mathrm{2}\right)\right)\:\Rightarrow \\ $$$${I}\:=\:\frac{\pi}{\mathrm{2}}\:−\mathrm{2}{arctan}\left(\mathrm{2}\right)+\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{5}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:+\frac{.\pi}{\mathrm{4}}\:−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$${I}\:=\pi\:−\mathrm{2}{arctan}\left(\mathrm{2}\right)\:+\frac{{ln}\left(\mathrm{5}\right)−{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\:. \\ $$
Commented by maxmathsup by imad last updated on 18/May/19
$${error}\:{at}\:{final}\:{line}\:{I}\:=\:\pi\:−\mathrm{2}{arctan}\left(\mathrm{2}\right)\:+\frac{{ln}\left(\mathrm{5}\right)}{\mathrm{2}}\:−{ln}\left(\mathrm{2}\right)\:. \\ $$
Answered by tanmay last updated on 07/May/19
![∫_0 ^1 tan^(−1) (1+x+x^2 )dx ∫_0 ^1 cot^(−1) ((1/(1+x(x+1))))dx x+1=tana x=tanb (1/(1+x(x+1)))=((tana−tanb)/(1+tanatanb))=tan(a−b)=cot[(π/2)−(a−b)] cot^(−1) [cot{(π/2)−(a−b)}] =(π/2)−a+b =(π/2)−tan^− (x+1)+tan^(−1) x ∫_0 ^1 (π/2)dx−∫_0 ^1 tan^(−1) (x+1)dx+∫_0 ^1 tan^(−1) xdx now ∫_0 ^1 (π/2)dx=(π/2) ∫_0 ^1 tan^(−1) (x+1)dx ★∫tan^(−1) (x+1)dx =tan^(−1) (x+1)×x−∫(x/(1+(1+x)^2 ))dx =xtan^(−1) (x+1)−(1/2)∫((2x+2−2)/(x^2 +2x+2)) =xtan^(−1) (x+1)−(1/2)∫((d(x^2 +2x+2))/(x^2 +2x+2))+∫(dx/(1+(x+1)^2 )) =xtan^(−1) (x+1)−(1/2)ln(x^2 +2x+2)+tan^(−1) (x+1) so ∣xtan^(−1) (x+1)−(1/2)ln(x^2 +2x+2)+tan^(−1) (x+1)∣_0 ^1 =■tan^(−1) (2)−(1/2)ln(5)+tan^− (2)+(1/2)ln2−(π/4)■ ∫_0 ^1 tan^(−1) xdx ∫tan^(−1) xdx =xtan^(−1) x−∫(x/(1+x^2 ))dx =xtan^(−1) x−(1/2)ln(1+x^2 ) so∣xtan^(−1) x−(1/2)ln(1+x^2 )∣_0 ^1 =tan^(−1) (1)−(1/2)ln2 =(π/4)−(1/2)ln2 so answer is ∫_0 ^1 (π/2)dx+∫_0 ^1 tan^(−1) xdx−∫_0 ^1 tan^(−1) (x+1)dx =(π/2)+(π/4)−(1/2)ln2−2tan^(−1) (2)+(1/2)ln5−(1/2)ln2+(π/4) =π−ln2+(1/2)ln5−2tan^(−1) (2) pls go through...](https://www.tinkutara.com/question/Q59311.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} \left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right){dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {cot}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}\left({x}+\mathrm{1}\right)}\right){dx} \\ $$$${x}+\mathrm{1}={tana}\:\:\:{x}={tanb} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{x}\left({x}+\mathrm{1}\right)}=\frac{{tana}−{tanb}}{\mathrm{1}+{tanatanb}}={tan}\left({a}−{b}\right)={cot}\left[\frac{\pi}{\mathrm{2}}−\left({a}−{b}\right)\right] \\ $$$${cot}^{−\mathrm{1}} \left[{cot}\left\{\frac{\pi}{\mathrm{2}}−\left({a}−{b}\right)\right\}\right] \\ $$$$=\frac{\pi}{\mathrm{2}}−{a}+{b} \\ $$$$=\frac{\pi}{\mathrm{2}}−{tan}^{−} \left({x}+\mathrm{1}\right)+{tan}^{−\mathrm{1}} {x} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\pi}{\mathrm{2}}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} \left({x}+\mathrm{1}\right){dx}+\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} {xdx} \\ $$$${now}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\pi}{\mathrm{2}}{dx}=\frac{\pi}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} \left({x}+\mathrm{1}\right){dx} \\ $$$$\bigstar\int{tan}^{−\mathrm{1}} \left({x}+\mathrm{1}\right){dx} \\ $$$$={tan}^{−\mathrm{1}} \left({x}+\mathrm{1}\right)×{x}−\int\frac{{x}}{\mathrm{1}+\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx} \\ $$$$={xtan}^{−\mathrm{1}} \left({x}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\mathrm{2}−\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}} \\ $$$$={xtan}^{−\mathrm{1}} \left({x}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}+\int\frac{{dx}}{\mathrm{1}+\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$={xtan}^{−\mathrm{1}} \left({x}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)+{tan}^{−\mathrm{1}} \left({x}+\mathrm{1}\right) \\ $$$${so}\:\mid{xtan}^{−\mathrm{1}} \left({x}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)+{tan}^{−\mathrm{1}} \left({x}+\mathrm{1}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\blacksquare{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{5}\right)+{tan}^{−} \left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}−\frac{\pi}{\mathrm{4}}\blacksquare \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} {xdx} \\ $$$$\int{tan}^{−\mathrm{1}} {xdx} \\ $$$$={xtan}^{−\mathrm{1}} {x}−\int\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$={xtan}^{−\mathrm{1}} {x}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right) \\ $$$$\boldsymbol{{so}}\mid{xtan}^{−\mathrm{1}} {x}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$={tan}^{−\mathrm{1}} \left(\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2} \\ $$$$=\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{answer}}\:\boldsymbol{{is}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\pi}{\mathrm{2}}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} {xdx}−\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} \left({x}+\mathrm{1}\right){dx} \\ $$$$=\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}−\mathrm{2}{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{5}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}+\frac{\pi}{\mathrm{4}} \\ $$$$=\pi−{ln}\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{5}−\mathrm{2}{tan}^{−\mathrm{1}} \left(\mathrm{2}\right) \\ $$$${pls}\:{go}\:{through}… \\ $$