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x-2-x-x-6-1-dx-




Question Number 150007 by mathdanisur last updated on 08/Aug/21
∫  ((x^2  + x)/(x^6  + 1)) dx = ?
$$\int\:\:\frac{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}}{\mathrm{x}^{\mathrm{6}} \:+\:\mathrm{1}}\:\mathrm{dx}\:=\:? \\ $$
Answered by Ar Brandon last updated on 08/Aug/21
I=∫((x^2 +x)/(x^6 +1))dx=∫(x^2 /(x^6 +1))dx+∫(x/(x^6 +1))dx    =(1/3)∫((d(x^3 ))/((x^3 )^2 +1))+(1/2)∫((d(x^2 ))/((x^2 )^3 +1))    =((tan^(−1) (x^3 ))/3)+(1/2)∫((d(x^2 ))/((x^2 +1)(x^4 −x^2 +1)))    =((tan^(−1) (x^3 ))/3)+(1/2)∫((1/3)∙(1/(x^2 +1))−(1/3)∙((x^2 −2)/(x^4 −x^2 +1)))dx^2     =((tan^(−1) (x^3 ))/3)+(1/6)ln(x^2 +1)−(1/(12))ln(x^4 −x^2 +1)+(1/(2(√3)))tan^(−1) (((2x^2 −1)/( (√3))))+C    =((tan^(−1) (x^3 ))/3)+(1/(12))ln∣((x^4 +2x^2 +1)/(x^4 −x^2 +1))∣+ ((√3)/6)tan^(−1) (((2x^2 −1)/( (√3))))+C
$${I}=\int\frac{{x}^{\mathrm{2}} +{x}}{{x}^{\mathrm{6}} +\mathrm{1}}{dx}=\int\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{6}} +\mathrm{1}}{dx}+\int\frac{{x}}{{x}^{\mathrm{6}} +\mathrm{1}}{dx} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{d}\left({x}^{\mathrm{3}} \right)}{\left({x}^{\mathrm{3}} \right)^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} \right)^{\mathrm{3}} +\mathrm{1}} \\ $$$$\:\:=\frac{\mathrm{tan}^{−\mathrm{1}} \left({x}^{\mathrm{3}} \right)}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\:\:=\frac{\mathrm{tan}^{−\mathrm{1}} \left({x}^{\mathrm{3}} \right)}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{1}}{\mathrm{3}}\centerdot\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}}\centerdot\frac{{x}^{\mathrm{2}} −\mathrm{2}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}}\right){dx}^{\mathrm{2}} \\ $$$$\:\:=\frac{\mathrm{tan}^{−\mathrm{1}} \left({x}^{\mathrm{3}} \right)}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{12}}\mathrm{ln}\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{C} \\ $$$$\:\:=\frac{\mathrm{tan}^{−\mathrm{1}} \left({x}^{\mathrm{3}} \right)}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{12}}\mathrm{ln}\mid\frac{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}}\mid+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{6}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{C} \\ $$
Commented by mathdanisur last updated on 08/Aug/21
Thank you Ser
$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

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