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Question Number 32951 by math1967 last updated on 07/Apr/18

Evaluate ∫((x^4 +1)/(x^6 +1))dx [W.B.H.S 2018]

$${Evaluate} \\ $$$$\int\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{6}} +\mathrm{1}}{dx}\:\:\:\:\:\left[{W}.{B}.{H}.{S}\:\mathrm{2018}\right] \\ $$

Commented by abdo imad last updated on 09/Apr/18

yes i have commited a error becaus i don t give attention to limits(bornes) thank for this remark... i have deleted the post..

$${yes}\:{i}\:{have}\:{commited}\:{a}\:{error}\:\:{becaus}\:{i}\:{don}\:{t}\:{give}\: \\ $$$${attention}\:{to}\:{limits}\left({bornes}\right)\:{thank}\:{for}\:{this}\:{remark}... \\ $$$${i}\:{have}\:{deleted}\:{the}\:{post}.. \\ $$

Answered by Joel578 last updated on 07/Apr/18

I = ∫ ((x^4 + 1)/(x^6 + 1)) dx = ∫ ((x^4 + 1 + x^2 − x^2 )/((x^2 + 1)(x^4 − x^2 + 1))) dx = ∫ ((x^4 − x^2 + 1)/((x^2 + 1)(x^4 − x^2 + 1))) dx + ∫ (x^2 /((x^2 + 1)(x^4 − x^2 + 1))) dx = ∫ (1/(x^2 + 1)) dx + ∫ (x^2 /(x^6 + 1)) dx I_2 = ∫ (x^2 /(x^6 + 1)) dx (u = x^3 → du = 3x^2 dx) = ∫ (x^2 /(u^2 + 1)) ((du/(3x^2 ))) = (1/3) ∫ (du/(u^2 + 1)) = (1/3)tan^(−1) (u) + C I = ∫ (1/(x^2 + 1)) dx + ∫ (x^2 /(x^6 + 1)) dx = tan^(−1) (x) + (1/3)tan^(−1) (u) + C = tan^(−1) (x) + (1/3)tan^(−1) (x^3 ) + C

$${I}\:=\:\int\:\frac{{x}^{\mathrm{4}} \:+\:\mathrm{1}}{{x}^{\mathrm{6}} \:+\:\mathrm{1}}\:{dx}\:=\:\int\:\frac{{x}^{\mathrm{4}} \:+\:\mathrm{1}\:+\:{x}^{\mathrm{2}} \:−\:{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{4}} \:−\:{x}^{\mathrm{2}} \:+\:\mathrm{1}\right)}\:{dx} \\ $$$$\:\:\:\:=\:\int\:\frac{{x}^{\mathrm{4}} \:−\:{x}^{\mathrm{2}} \:+\:\mathrm{1}}{\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{4}} \:−\:{x}^{\mathrm{2}} \:+\:\mathrm{1}\right)}\:\:{dx}\:+\:\int\:\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{4}} \:−\:{x}^{\mathrm{2}} \:+\:\mathrm{1}\right)}\:{dx} \\ $$$$\:\:\:\:=\:\int\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\:\mathrm{1}}\:{dx}\:+\:\int\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{6}} \:+\:\mathrm{1}}\:{dx} \\ $$$$\: \\ $$$${I}_{\mathrm{2}} \:=\:\int\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{6}} \:+\:\mathrm{1}}\:{dx}\:\:\:\left({u}\:=\:{x}^{\mathrm{3}} \:\:\rightarrow\:\:{du}\:=\:\mathrm{3}{x}^{\mathrm{2}} \:{dx}\right) \\ $$$$\:\:\:\:\:=\:\int\:\frac{{x}^{\mathrm{2}} }{{u}^{\mathrm{2}} \:+\:\mathrm{1}}\:\left(\frac{{du}}{\mathrm{3}{x}^{\mathrm{2}} }\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\int\:\frac{{du}}{{u}^{\mathrm{2}} \:+\:\mathrm{1}} \\ $$$$\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \left({u}\right)\:+\:{C} \\ $$$$ \\ $$$${I}\:=\:\int\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\:\mathrm{1}}\:{dx}\:+\:\int\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{6}} \:+\:\mathrm{1}}\:{dx} \\ $$$$\:\:\:=\:\mathrm{tan}^{−\mathrm{1}} \:\left({x}\right)\:+\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \:\left({u}\right)\:+\:{C} \\ $$$$\:\:\:=\:\mathrm{tan}^{−\mathrm{1}} \:\left({x}\right)\:+\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \:\left({x}^{\mathrm{3}} \right)\:+\:{C} \\ $$

Answered by MJS last updated on 07/Apr/18

lol! I did it the hard way... x^6 +1=(x^4 −x^2 +1)(x^2 +1)= =(x^2 −(√3)x+1)(x^2 +(√3)x+1)(x^2 +1) ((x^4 +1)/(x^6 +1))=(N_1 /(x^2 −(√3)x+1))+(N_2 /(x^2 +(√3)+1))+(N_3 /(x^2 +1 )) N_1 =N_2 =(1/6); N_3 =(2/3) ∫((x^4 +1)/(x^6 +1))dx=∫(1/(6(x^2 −(√3)x+1)))dx+∫(1/(6(x^2 +(√3)x+1)))dx+∫(2/(3(x^2 +1)))dx= [∫(1/(ax^2 +bx+c))dx=(2/(√(−b^2 +4ac)))tan^(−1) ((2ax+b)/(√(−b^2 +4ac))); b^2 −4ac<0] =(1/3)(tan^(−1) (2x−(√3))+tan^(−1) (2x+(√3))+2tan^(−1) (x))+C

$$\mathrm{lol}!\:\mathrm{I}\:\mathrm{did}\:\mathrm{it}\:\mathrm{the}\:\mathrm{hard}\:\mathrm{way}... \\ $$$${x}^{\mathrm{6}} +\mathrm{1}=\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)= \\ $$$$=\left({x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{6}} +\mathrm{1}}=\frac{{N}_{\mathrm{1}} }{{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}}+\frac{{N}_{\mathrm{2}} }{{x}^{\mathrm{2}} +\sqrt{\mathrm{3}}+\mathrm{1}}+\frac{{N}_{\mathrm{3}} }{{x}^{\mathrm{2}} +\mathrm{1}\:} \\ $$$${N}_{\mathrm{1}} ={N}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{6}};\:{N}_{\mathrm{3}} =\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\int\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{6}} +\mathrm{1}}{dx}=\int\frac{\mathrm{1}}{\mathrm{6}\left({x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)}{dx}+\int\frac{\mathrm{1}}{\mathrm{6}\left({x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}\right)}{dx}+\int\frac{\mathrm{2}}{\mathrm{3}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}= \\ $$$$\:\:\:\:\:\left[\int\frac{\mathrm{1}}{{ax}^{\mathrm{2}} +{bx}+{c}}{dx}=\frac{\mathrm{2}}{\sqrt{−{b}^{\mathrm{2}} +\mathrm{4}{ac}}}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{ax}+{b}}{\sqrt{−{b}^{\mathrm{2}} +\mathrm{4}{ac}}};\:{b}^{\mathrm{2}} −\mathrm{4}{ac}<\mathrm{0}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{x}−\sqrt{\mathrm{3}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{x}+\sqrt{\mathrm{3}}\right)+\mathrm{2tan}^{−\mathrm{1}} \left({x}\right)\right)+{C} \\ $$

Commented by Joel578 last updated on 08/Apr/18

Great work Sir

$$\mathrm{Great}\:\mathrm{work}\:\mathrm{Sir} \\ $$

Commented by MJS last updated on 08/Apr/18

thanks, but I like your solution more than mine

$$\mathrm{thanks},\:\mathrm{but}\:\mathrm{I}\:\mathrm{like}\:\mathrm{your}\:\mathrm{solution} \\ $$$$\mathrm{more}\:\mathrm{than}\:\mathrm{mine} \\ $$

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