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1-dx-x-4-2x-2-9-2-arc-tan-x-3-arc-tan-x-3-arc-tan-1-5-

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Question Number 121500 by bramlexs22 last updated on 08/Nov/20
(1) ∫ (dx/(x^4 +2x^2 +9)) ? (2) arc tan ((x/3))−arc tan ((x/3))= arc tan ((1/5))
$$\:\left(\mathrm{1}\right)\:\int\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} +\mathrm{2x}^{\mathrm{2}} +\mathrm{9}}\:? \\ $$$$\left(\mathrm{2}\right)\:\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{3}}\right)−\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{3}}\right)=\:\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{1}}{\mathrm{5}}\right) \\ $$
Commented by benjo_mathlover last updated on 08/Nov/20
i think it should be (2) tan^(−1) ((x/3))−tan^(−1) ((x/2))= tan^(−1) ((1/5)) let x = 6ψ ⇒ tan^(−1) (2ψ)−tan^(−1) (3ψ)=tan^(−1) ((1/5)) ⇔ ((2ψ−3ψ)/(1+6ψ^2 )) = (1/5) ; 6ψ^2 +5ψ+1=0 (2ψ+1)(3ψ+1)=0 → { (((1/6)x=−(1/2);x =−3)),(((1/6)x=−(1/3); x=−2)) :}
$$\mathrm{i}\:\mathrm{think}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\: \\ $$$$\left(\mathrm{2}\right)\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{3}}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right) \\ $$$$\mathrm{let}\:\mathrm{x}\:=\:\mathrm{6}\psi\:\Rightarrow\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\psi\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{3}\psi\right)=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right) \\ $$$$\Leftrightarrow\:\frac{\mathrm{2}\psi−\mathrm{3}\psi}{\mathrm{1}+\mathrm{6}\psi^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{5}}\:;\:\mathrm{6}\psi^{\mathrm{2}} +\mathrm{5}\psi+\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{2}\psi+\mathrm{1}\right)\left(\mathrm{3}\psi+\mathrm{1}\right)=\mathrm{0}\:\rightarrow\begin{cases}{\frac{\mathrm{1}}{\mathrm{6}}\mathrm{x}=−\frac{\mathrm{1}}{\mathrm{2}};\mathrm{x}\:=−\mathrm{3}}\\{\frac{\mathrm{1}}{\mathrm{6}}\mathrm{x}=−\frac{\mathrm{1}}{\mathrm{3}};\:\mathrm{x}=−\mathrm{2}}\end{cases} \\ $$
Answered by Ar Brandon last updated on 09/Nov/20
1\I=∫(dx/(x^4 +2x^2 +9))=(1/6)∫(((x^2 +3)−(x^2 −3))/(x^4 +2x^2 +9))dx 6I=∫((x^2 +3)/(x^4 +2x^2 +9))dx−∫((x^2 −3)/(x^4 +2x^2 +9))dx =∫((1+(3/x^2 ))/(x^2 +2+(9/x^2 )))dx−∫((1−(3/x^2 ))/(x^2 +2+(9/x^2 )))dx =∫((1+(3/x^2 ))/((x−(3/x))^2 +8))dx−∫((1−(3/x^2 ))/((x+(3/x))^2 −4))dx =∫(du/(u^2 +8))−∫(dv/(v^2 −4))=(1/(2(√2)))Arctan((u/(2(√2))))+(1/2)Arctanh((v/2))+C
$$\mathrm{1}\backslash\mathcal{I}=\int\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} +\mathrm{2x}^{\mathrm{2}} +\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3}\right)−\left(\mathrm{x}^{\mathrm{2}} −\mathrm{3}\right)}{\mathrm{x}^{\mathrm{4}} +\mathrm{2x}^{\mathrm{2}} +\mathrm{9}}\mathrm{dx} \\ $$$$\:\:\:\mathrm{6}\mathcal{I}=\int\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{3}}{\mathrm{x}^{\mathrm{4}} +\mathrm{2x}^{\mathrm{2}} +\mathrm{9}}\mathrm{dx}−\int\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{3}}{\mathrm{x}^{\mathrm{4}} +\mathrm{2x}^{\mathrm{2}} +\mathrm{9}}\mathrm{dx} \\ $$$$\:\:\:\:\:\:\:\:\:=\int\frac{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{9}}{\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}−\int\frac{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{9}}{\mathrm{x}^{\mathrm{2}} }}\mathrm{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\int\frac{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }}{\left(\mathrm{x}−\frac{\mathrm{3}}{\mathrm{x}}\right)^{\mathrm{2}} +\mathrm{8}}\mathrm{dx}−\int\frac{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }}{\left(\mathrm{x}+\frac{\mathrm{3}}{\mathrm{x}}\right)^{\mathrm{2}} −\mathrm{4}}\mathrm{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\int\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} +\mathrm{8}}−\int\frac{\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} −\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{Arctan}\left(\frac{\mathrm{u}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Arctanh}\left(\frac{\mathrm{v}}{\mathrm{2}}\right)+\mathcal{C} \\ $$
Answered by MJS_new last updated on 09/Nov/20
∫(dx/(x^4 +2x^2 +9))=∫(dx/((x^2 −2x+3)(x^2 +2x+3)))= =(1/(12))∫(((x+2)/(x^2 +2x+3))−((x−2)/(x^2 −2x+3)))dx= =(1/(12))∫(dx/(x^2 +2x+3))+(1/(24))∫((2x+2)/(x^2 +2x+3))dx+ +(1/(12))∫(dx/(x^2 −2x+3))−(1/(24))∫((2x−2)/(x^2 −2x+3))dx= =((√2)/(24))arctan (((√2)(x+1))/2) +(1/(24))ln (x^2 +2x+3) + +((√2)/(24))arctan (((√2)(x−1))/2) −(1/(24))ln (x^2 −2x+3) +C
$$\int\frac{{dx}}{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{9}}=\int\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\int\left(\frac{{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}−\frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}}\right){dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{24}}\int\frac{\mathrm{2}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}{dx}+ \\ $$$$+\frac{\mathrm{1}}{\mathrm{12}}\int\frac{{dx}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{24}}\int\frac{\mathrm{2}{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}}{dx}= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{24}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{2}}\left({x}+\mathrm{1}\right)}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{24}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)\:+ \\ $$$$+\frac{\sqrt{\mathrm{2}}}{\mathrm{24}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{2}}\left({x}−\mathrm{1}\right)}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{24}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}\right)\:+{C} \\ $$
Answered by TANMAY PANACEA last updated on 09/Nov/20
1)∫((1/x^2 )/(x^2 +(9/x^2 )+2))dx (1/6)∫(((1+(3/x^2 ))−(1−(3/x^2 )))/(x^2 +(9/x^2 )+2))dx (1/6)∫((d(x−(3/x)))/((x−(3/(x )))^2 +8)) −(1/6)∫((d(x+(3/x)))/((x+(3/x))^2 −4)) (1/6)×(1/( (√8)))tan^(−1) (((x−(3/x))/( (√8))))−(1/6)×(1/(2×2))ln(((x−(3/x)−2)/(x−(3/x)+2)))+c
$$\left.\mathrm{1}\right)\int\frac{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} +\frac{\mathrm{9}}{{x}^{\mathrm{2}} }+\mathrm{2}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\left(\mathrm{1}+\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\right)−\left(\mathrm{1}−\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{2}} +\frac{\mathrm{9}}{{x}^{\mathrm{2}} }+\mathrm{2}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\int\frac{{d}\left({x}−\frac{\mathrm{3}}{{x}}\right)}{\left({x}−\frac{\mathrm{3}}{{x}\:}\right)^{\mathrm{2}} +\mathrm{8}}\:−\frac{\mathrm{1}}{\mathrm{6}}\int\frac{{d}\left({x}+\frac{\mathrm{3}}{{x}}\right)}{\left({x}+\frac{\mathrm{3}}{{x}}\right)^{\mathrm{2}} −\mathrm{4}} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{8}}}{tan}^{−\mathrm{1}} \left(\frac{{x}−\frac{\mathrm{3}}{{x}}}{\:\sqrt{\mathrm{8}}}\right)−\frac{\mathrm{1}}{\mathrm{6}}×\frac{\mathrm{1}}{\mathrm{2}×\mathrm{2}}{ln}\left(\frac{{x}−\frac{\mathrm{3}}{{x}}−\mathrm{2}}{{x}−\frac{\mathrm{3}}{{x}}+\mathrm{2}}\right)+{c} \\ $$

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