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Question Number 129057 by pipin last updated on 12/Jan/21

∫_1 ^∞ (1/(1+x^3 ))dx = ...

$$\: \\ $$$$\:\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{3}} }\mathrm{dx}\:=\:... \\ $$

Answered by MJS_new last updated on 12/Jan/21

∫(dx/(x^3 +1))=∫(dx/((x+1)(x^2 −x+1)))= =(1/3)∫(dx/(x+1))−(1/3)∫((x−2)/(x^2 −x+1))dx= =(1/3)∫(dx/(x+1))−(1/6)∫((2x−1)/(x^2 −x+1))dx+(1/2)∫(dx/(x^2 −x+1))= =(1/3)ln ∣x+1∣ −(1/6)ln (x^2 −x+1) +((√3)/3)arctan (((√3)(2x−1))/3) +C ⇒ answer is ((π(√3))/9)−((ln 2)/3)

$$\int\frac{{dx}}{{x}^{\mathrm{3}} +\mathrm{1}}=\int\frac{{dx}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\mid{x}+\mathrm{1}\mid\:−\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{x}−\mathrm{1}\right)}{\mathrm{3}}\:+{C} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{9}}−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{3}} \\ $$

Answered by Dwaipayan Shikari last updated on 12/Jan/21

∫_0 ^∞ (1/(1+x^3 ))−(1/3)∫_0 ^1 (1/(1+x))−((x−2)/(x^2 −x+1))dx =(1/3)∫_0 ^∞ (u^(−(2/3)) /(1+u))du −(1/3)∫_0 ^1 (1/(1+x))+(1/6)∫_0 ^1 ((2x−1)/(x^2 −x+1))−(1/2)∫(1/(x^2 −x+1)) =(1/3).(π/(sin((π/3))))−(1/3)log(2)−(1/( (√3)))[tan^(−1) ((2x−1)/( (√3)))]_0 ^1 =((2π)/(3(√3)))−(1/3)log(2)−(π/( 3(√3)))=(π/( 3(√3)))−(1/3)log(2)

$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{x}}−\frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{−\frac{\mathrm{2}}{\mathrm{3}}} }{\mathrm{1}+{u}}{du}\:−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}.\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{3}}\right)}−\frac{\mathrm{1}}{\mathrm{3}}{log}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left[{tan}^{−\mathrm{1}} \frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}−\frac{\mathrm{1}}{\mathrm{3}}{log}\left(\mathrm{2}\right)−\frac{\pi}{\:\mathrm{3}\sqrt{\mathrm{3}}}=\frac{\pi}{\:\mathrm{3}\sqrt{\mathrm{3}}}−\frac{\mathrm{1}}{\mathrm{3}}{log}\left(\mathrm{2}\right) \\ $$

Commented by MJS_new last updated on 12/Jan/21

I had misread the lower border...

$$\mathrm{I}\:\mathrm{had}\:\mathrm{misread}\:\mathrm{the}\:\mathrm{lower}\:\mathrm{border}... \\ $$

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