Question Number 46856 by maxmathsup by imad last updated on 01/Nov/18
$${find}\:{f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{2}} \:{arctan}\left(\mathrm{1}+{tx}\right){dx}\: \\ $$
Commented by maxmathsup by imad last updated on 02/Nov/18
![changement tx =u give f(t) =∫_0 ^t (u^2 /t^2 ) arctan(1+u)(du/t) = (1/t^3 ) ∫_0 ^t u^2 arctan(1+u)du (we suppose u≠0) but ∫_0 ^t u^2 arctan(1+u)du =_(1+u =α) ∫_1 ^(1+t) (α−1)^2 arctanα dα =[(1/3)(α−1)^3 arctanα]_1 ^(1+t) −(1/3)∫_1 ^(1+t) (α−1)^3 (dα/(1+α^2 )) =(1/3){t^3 arctan(1+t)} −(1/3) ∫_1 ^(1+t) ((α^3 −3α^2 +3α −1)/(1+α^2 ))dα ∫_1 ^(1+t) ((α^3 −3α^2 +3α −1)/(1+α^2 ))dα =∫_1 ^(1+t) ((α(α^2 +1)+2α −3α^2 −1)/(1+α^2 )) dα =∫_1 ^(1+t) α dα +∫_1 ^(1+t) ((2α)/(1+α^2 )) dα − ∫_1 ^(1+α) ((3α^2 +1)/(1+α^2 )) dα =[(α^2 /2)]_1 ^(1+t) +[ln(1+α^2 )]_1 ^(1+t) −3 ∫_1 ^(1+t) ((1+α^2 −1)/(1+α^2 )) dα −[arctanα]_1 ^(1+t) =(1/2){(1+t)^2 −1} +ln(1+(1+t)^2 )−ln(2)−3 t +[2 arctanα]_1 ^(1+t) =(1/2){ t^2 +2t}+ln(t^2 +2t +2)−ln(2)−3t +2{ artan (1+t)−(π/2)}.](https://www.tinkutara.com/question/Q46949.png)
$${changement}\:{tx}\:={u}\:{give}\:{f}\left({t}\right)\:=\int_{\mathrm{0}} ^{{t}} \:\frac{{u}^{\mathrm{2}} }{{t}^{\mathrm{2}} }\:{arctan}\left(\mathrm{1}+{u}\right)\frac{{du}}{{t}} \\ $$$$=\:\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\:\int_{\mathrm{0}} ^{{t}} \:{u}^{\mathrm{2}} \:{arctan}\left(\mathrm{1}+{u}\right){du}\:\:\left({we}\:{suppose}\:{u}\neq\mathrm{0}\right)\:{but} \\ $$$$\int_{\mathrm{0}} ^{{t}} \:{u}^{\mathrm{2}} \:{arctan}\left(\mathrm{1}+{u}\right){du}\:=_{\mathrm{1}+{u}\:=\alpha} \:\:\int_{\mathrm{1}} ^{\mathrm{1}+{t}} \:\left(\alpha−\mathrm{1}\right)^{\mathrm{2}} \:{arctan}\alpha\:{d}\alpha \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{3}}\left(\alpha−\mathrm{1}\right)^{\mathrm{3}} \:{arctan}\alpha\right]_{\mathrm{1}} ^{\mathrm{1}+{t}} \:−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{1}} ^{\mathrm{1}+{t}} \:\:\left(\alpha−\mathrm{1}\right)^{\mathrm{3}} \:\frac{{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left\{{t}^{\mathrm{3}} \:{arctan}\left(\mathrm{1}+{t}\right)\right\}\:−\frac{\mathrm{1}}{\mathrm{3}}\:\int_{\mathrm{1}} ^{\mathrm{1}+{t}} \:\frac{\alpha^{\mathrm{3}} \:−\mathrm{3}\alpha^{\mathrm{2}} \:+\mathrm{3}\alpha\:−\mathrm{1}}{\mathrm{1}+\alpha^{\mathrm{2}} }{d}\alpha \\ $$$$\int_{\mathrm{1}} ^{\mathrm{1}+{t}} \:\frac{\alpha^{\mathrm{3}} \:−\mathrm{3}\alpha^{\mathrm{2}} \:+\mathrm{3}\alpha\:−\mathrm{1}}{\mathrm{1}+\alpha^{\mathrm{2}} }{d}\alpha\:=\int_{\mathrm{1}} ^{\mathrm{1}+{t}} \:\:\frac{\alpha\left(\alpha^{\mathrm{2}} \:+\mathrm{1}\right)+\mathrm{2}\alpha\:−\mathrm{3}\alpha^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+\alpha^{\mathrm{2}} }\:{d}\alpha \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{1}+{t}} \:\alpha\:{d}\alpha\:\:+\int_{\mathrm{1}} ^{\mathrm{1}+{t}} \:\:\frac{\mathrm{2}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} }\:{d}\alpha\:−\:\int_{\mathrm{1}} ^{\mathrm{1}+\alpha} \:\frac{\mathrm{3}\alpha^{\mathrm{2}} \:+\mathrm{1}}{\mathrm{1}+\alpha^{\mathrm{2}} }\:{d}\alpha \\ $$$$=\left[\frac{\alpha^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{1}} ^{\mathrm{1}+{t}} \:+\left[{ln}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)\right]_{\mathrm{1}} ^{\mathrm{1}+{t}} \:\:−\mathrm{3}\:\:\int_{\mathrm{1}} ^{\mathrm{1}+{t}} \:\frac{\mathrm{1}+\alpha^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+\alpha^{\mathrm{2}} }\:{d}\alpha\:−\left[{arctan}\alpha\right]_{\mathrm{1}} ^{\mathrm{1}+{t}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} −\mathrm{1}\right\}\:+{ln}\left(\mathrm{1}+\left(\mathrm{1}+{t}\right)^{\mathrm{2}} \right)−{ln}\left(\mathrm{2}\right)−\mathrm{3}\:{t}\:\:\:+\left[\mathrm{2}\:{arctan}\alpha\right]_{\mathrm{1}} ^{\mathrm{1}+{t}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{t}^{\mathrm{2}} \:+\mathrm{2}{t}\right\}+{ln}\left({t}^{\mathrm{2}} \:+\mathrm{2}{t}\:+\mathrm{2}\right)−{ln}\left(\mathrm{2}\right)−\mathrm{3}{t}\:\:+\mathrm{2}\left\{\:{artan}\:\left(\mathrm{1}+{t}\right)−\frac{\pi}{\mathrm{2}}\right\}. \\ $$