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Question Number 55995 by maxmathsup by imad last updated on 07/Mar/19
find I =∫ arctan(1−x)dx and J =∫ actan(1+x) dx
$${find}\:{I}\:=\int\:\:{arctan}\left(\mathrm{1}−{x}\right){dx}\:\:{and}\:{J}\:=\int\:{actan}\left(\mathrm{1}+{x}\right)\:{dx} \\ $$
Commented by maxmathsup by imad last updated on 08/Mar/19
by psrts I =x arctan(1−x)−∫ x ((−1)/(1+(1−x)^2 ))dx =xarctan(1−x) +∫ (x/(1+x^2 −2x+1))dx but ∫ (x/(x^2 −2x+2))dx =(1/2)∫ ((2x−2+2)/(x^2 −2x+2))dx =(1/2)ln∣x^2 −2x+2∣ +∫ (dx/(x^2 −2x+2)) +c ∫ (dx/(x^2 −2x+2))dx =∫ (dx/((x−1)^2 +1)) =_(x−1 =u) ∫ (du/(1+u^2 )) =arctan(u) =arctan(x−1) ⇒I = xarctan(1−x)+(1/2)ln∣x^2 −2x+2∣ +arctan(x−1) +c =(x−1) arctan(1−x) +ln(√(x^2 −2x+2))+c .also J =∫ arctan(1+x)dx =_(x=−t) − ∫ arctan(1−t)dt =−(t−1)arctan(1−t)+ln(√(t^2 −2t +2))) +c =(x+1)arctan(1+x)+ln(√(x^2 +2x+2)) +c .
$${by}\:{psrts}\:{I}\:={x}\:{arctan}\left(\mathrm{1}−{x}\right)−\int\:{x}\:\frac{−\mathrm{1}}{\mathrm{1}+\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{dx} \\ $$$$={xarctan}\left(\mathrm{1}−{x}\right)\:+\int\:\:\:\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}{dx}\:\:{but} \\ $$$$\int\:\:\:\:\frac{{x}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}}{dx}\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{\mathrm{2}{x}−\mathrm{2}+\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}}{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\mid\:+\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}}\:+{c} \\ $$$$\int\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}}{dx}\:=\int\:\:\:\frac{{dx}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}}\:=_{{x}−\mathrm{1}\:={u}} \:\:\:\int\:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:={arctan}\left({u}\right)\:={arctan}\left({x}−\mathrm{1}\right) \\ $$$$\Rightarrow{I}\:=\:{xarctan}\left(\mathrm{1}−{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\mid\:+{arctan}\left({x}−\mathrm{1}\right)\:+{c} \\ $$$$=\left({x}−\mathrm{1}\right)\:{arctan}\left(\mathrm{1}−{x}\right)\:+{ln}\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}}+{c}\:\:\:.{also}\: \\ $$$${J}\:=\int\:\:{arctan}\left(\mathrm{1}+{x}\right){dx}\:=_{{x}=−{t}} −\:\int\:{arctan}\left(\mathrm{1}−{t}\right){dt} \\ $$$$\left.=−\left({t}−\mathrm{1}\right){arctan}\left(\mathrm{1}−{t}\right)+{ln}\sqrt{{t}^{\mathrm{2}} −\mathrm{2}{t}\:+\mathrm{2}}\right)\:+{c} \\ $$$$=\left({x}+\mathrm{1}\right){arctan}\left(\mathrm{1}+{x}\right)+{ln}\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}\:+{c}\:. \\ $$
Answered by MJS last updated on 08/Mar/19
they are standard?! arctan (1−x) =−arctan (x−1) ∫arctan (x±a) dx= =(x±a)arctan (x±a)−(1/2)ln ((x±a)^2 +1) +C
$$\mathrm{they}\:\mathrm{are}\:\mathrm{standard}?! \\ $$$$\mathrm{arctan}\:\left(\mathrm{1}−{x}\right)\:=−\mathrm{arctan}\:\left({x}−\mathrm{1}\right) \\ $$$$\int\mathrm{arctan}\:\left({x}\pm{a}\right)\:{dx}= \\ $$$$=\left({x}\pm{a}\right)\mathrm{arctan}\:\left({x}\pm{a}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\left({x}\pm{a}\right)^{\mathrm{2}} +\mathrm{1}\right)\:+{C} \\ $$

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