Question Number 60901 by maxmathsup by imad last updated on 27/May/19
$${find}\:\int\:\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx}\: \\ $$
Answered by perlman last updated on 27/May/19
$${integrat}\:{by}\left[{part}\right. \\ $$$$={xarctan}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)+\int{x}×\left(\frac{\mathrm{2}{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }}\right){dx} \\ $$$$={xarctan}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)+\int\frac{\mathrm{2}{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$={xarctan}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)+\int\frac{\mathrm{2}{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} −{i}^{\mathrm{2}} } \\ $$$$={xarctg}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)+\int\frac{\mathrm{2}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}−{i}\right)\left({x}^{\mathrm{2}} +\mathrm{1}+{i}\right)}{dx} \\ $$$$={xarctg}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)+\int\frac{\mathrm{2}{x}\mathrm{2}}{\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{e}^{\mathrm{3}{i}\pi/\mathrm{4}} \right)\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{e}^{\mathrm{5}{i}\pi/\mathrm{4}} \right)}{dx} \\ $$$${xarctan}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right)+\int\frac{\mathrm{2}{x}^{\mathrm{2}\:\:} }{\left({x}−\mathrm{2}^{\mathrm{1}/\mathrm{4}} {e}^{\left(\mathrm{3}{i}\pi/\mathrm{8}\right)} \right)\left({x}+\mathrm{2}^{\mathrm{1}/\mathrm{4}} {e}^{\left(\mathrm{3}{i}\pi/\mathrm{8}\right)} \right)\left({x}−\mathrm{2}^{\mathrm{1}/\mathrm{4}} {e}^{\mathrm{5}{i}\pi/\mathrm{8}} \right)\left({x}+\mathrm{2}^{\mathrm{1}/\mathrm{4}} {e}_{} ^{\mathrm{5}{i}\pi/\mathrm{8}} \right)} \\ $$$$={xarctg}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)+\int\frac{\mathrm{2}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −\mathrm{2}×\mathrm{2}^{\mathrm{1}/\mathrm{4}} \left({cos}\left(\mathrm{3}\pi/\mathrm{8}\right){x}+\sqrt{\mathrm{2}}\right)\left({x}^{\mathrm{2}} +\mathrm{2}×\mathrm{2}^{\mathrm{1}/\mathrm{4}} {cos}\left(\mathrm{3}\pi/\mathrm{8}\right){x}+\sqrt{\mathrm{2}}\right)\right.} \\ $$$${put}\:\:\:{a}=\mathrm{2}×\mathrm{2}^{\mathrm{1}/\mathrm{4}} {cos}\left(\mathrm{3}\pi/\mathrm{8}\right) \\ $$$$={xarctan}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)+\int\frac{{sx}+{d}}{{x}^{\mathrm{2}} −{ax}+\sqrt{\mathrm{2}}}+\:\frac{{cx}+{v}}{{x}^{\mathrm{2}} +{ax}+\sqrt{\mathrm{2}}}{dx} \\ $$$${withe}\:{s}+{c}=\mathrm{0}.{d}+{v}=\mathrm{0}.{d}+{as}−{ac}+{v}=\mathrm{2}==>{s}−{c}=\frac{\mathrm{2}}{{a}}\:\:{s}+{c}=\mathrm{0} \\ $$$${s}=\frac{\mathrm{1}}{{a}}.{c}=−\frac{\mathrm{1}}{{a}}.{withe}\:{s}\sqrt{\mathrm{2}}+{da}−{av}+{c}\sqrt{\mathrm{2}}\:=\mathrm{0}===>{v}={d}=\mathrm{0} \\ $$$${our}\:{integral}\:{is} \\ $$$${xarctan}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)+\int\frac{{x}}{{a}\left({x}^{\mathrm{2}} −{ax}+\sqrt{\mathrm{2}}\right)}{dx}−\int\frac{{x}}{{a}\left({x}^{\mathrm{2}} +{ax}+\sqrt{\left.\mathrm{2}\right)}\right.}{dx} \\ $$$${lets}\:{find}\:\int\frac{{x}}{{a}\left({x}^{\mathrm{2}} −{ax}+\sqrt{\mathrm{2}}\right)}{dx}=\int\frac{\mathrm{2}{x}−{a}+{a}}{\mathrm{2}{a}\left({x}^{\mathrm{2}} −{ax}+\sqrt{\mathrm{2}}\right)}{dx}=\frac{{ln}\left({x}^{\mathrm{2}} −{ax}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}{a}}+\int\frac{\mathrm{1}}{\mathrm{2}\left({x}−{a}/\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}−{a}^{\mathrm{2}} /\mathrm{2}} \\ $$$${let}\:{b}^{\mathrm{2}\:} =\sqrt{\mathrm{2}}−{a}^{\mathrm{2}} /\mathrm{4} \\ $$$$=\frac{{ln}\left({x}^{\mathrm{2}} −{ax}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}{a}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\left({x}−{a}/\mathrm{2}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} }=\frac{{ln}\left({x}^{\mathrm{2}} −{ax}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}{a}}+\frac{{b}}{\mathrm{2}}{arctg}\left(\frac{\mathrm{2}{x}−{a}}{\mathrm{2}{b}}\right) \\ $$$${finaly}\:{we}\:{get}\:{xarctan}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)+\frac{{ln}\left({x}^{\mathrm{2}} −{ax}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}{a}}+\frac{{b}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{2}{x}−{a}}{\mathrm{2}{b}}\right) \\ $$$$+\frac{{ln}\left({x}^{\mathrm{2}} +{ax}+\sqrt{\mathrm{2}}\right)}{−\mathrm{2}{a}}+\frac{{b}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{2}{x}+{a}}{\mathrm{2}{b}}\right)+{c}\:\:\:\:\:{withe}\:{c}\:{constante} \\ $$$$ \\ $$