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Question Number 60901 by maxmathsup by imad last updated on 27/May/19

find ∫ arctan((1/(1+x^2 )))dx

$${find}\:\int\:\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx}\: \\ $$

Answered by perlman last updated on 27/May/19

integrat by[part =xarctan((1/(1+x^2 )))+∫x×(((2x)/((x^2 +1)^2 ))×(1/(1+(1/((1+x^2 )^2 )))))dx =xarctan((1/(1+x^2 )))+∫((2x^2 )/((1+x^2 )^2 +1))dx =xarctan((1/(1+x^2 )))+∫((2x^2 )/((1+x^2 )^2 −i^2 )) =xarctg((1/(1+x^2 )))+∫((2x^2 )/((x^2 +1−i)(x^2 +1+i)))dx =xarctg((1/(1+x^2 )))+∫((2x2)/((x^2 −(√2)e^(3iπ/4) )(x^2 −(√2)e^(5iπ/4) )))dx xarctan((1/(x^2 +1)))+∫((2x^(2 ) )/((x−2^(1/4) e^((3iπ/8)) )(x+2^(1/4) e^((3iπ/8)) )(x−2^(1/4) e^(5iπ/8) )(x+2^(1/4) e_ ^(5iπ/8) ))) =xarctg((1/(1+x^2 )))+∫((2x^2 )/((x^2 −2×2^(1/4) (cos(3π/8)x+(√2))(x^2 +2×2^(1/4) cos(3π/8)x+(√2)))) put a=2×2^(1/4) cos(3π/8) =xarctan((1/(1+x^2 )))+∫((sx+d)/(x^2 −ax+(√2)))+ ((cx+v)/(x^2 +ax+(√2)))dx withe s+c=0.d+v=0.d+as−ac+v=2==>s−c=(2/a) s+c=0 s=(1/a).c=−(1/a).withe s(√2)+da−av+c(√2) =0===>v=d=0 our integral is xarctan((1/(1+x^2 )))+∫(x/(a(x^2 −ax+(√2))))dx−∫(x/(a(x^2 +ax+(√(2)))))dx lets find ∫(x/(a(x^2 −ax+(√2))))dx=∫((2x−a+a)/(2a(x^2 −ax+(√2))))dx=((ln(x^2 −ax+(√2)))/(2a))+∫(1/(2(x−a/2)^2 +2(√2)−a^2 /2)) let b^(2 ) =(√2)−a^2 /4 =((ln(x^2 −ax+(√2)))/(2a))+(1/2)∫(1/((x−a/2)^2 +b^2 ))=((ln(x^2 −ax+(√2)))/(2a))+(b/2)arctg(((2x−a)/(2b))) finaly we get xarctan((1/(1+x^2 )))+((ln(x^2 −ax+(√2)))/(2a))+(b/2)arctan(((2x−a)/(2b))) +((ln(x^2 +ax+(√2)))/(−2a))+(b/2)arctan(((2x+a)/(2b)))+c withe c constante

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

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