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Question Number 158721 by HongKing last updated on 08/Nov/21
Find:  ∫_( 0) ^( ∞)  (1/((x^2  + x + 1)(1 + ax))) dx  ;  a>0
$$\mathrm{Find}: \\ $$$$\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{1}\right)\left(\mathrm{1}\:+\:\mathrm{ax}\right)}\:\mathrm{dx}\:\:;\:\:\mathrm{a}>\mathrm{0} \\ $$$$ \\ $$
Answered by ajfour last updated on 08/Nov/21
(1/a)∫(((Ax+B)/(x^2 +x+1))+(C/(x+(1/a))))dx  A+C=0  (A/a)+B+C=0  (B/a)+C=1  ⇒ (A/a^2 )+1+A−(A/a)=0  A=(1/((1/a)−1−(1/a^2 )))=(a^2 /(a−a^2 −1))  C=−A ; B=a(1+A)      B=((a(a−1))/(a(1−a)−1))  ∫(dx/(x+(1/a)))=ln (x+(1/a))+c_1   ∫(((2x+1)dx)/(x^2 +x+1))=ln (x^2 +x+1)+c_2   ∫(dx/((x+(1/2))^2 +(((√3)/2))^2 ))=          (2/( (√3)))tan^(−1) (((2x+1)/( (√3))))+c_3
$$\frac{\mathrm{1}}{{a}}\int\left(\frac{{Ax}+{B}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\frac{{C}}{{x}+\frac{\mathrm{1}}{{a}}}\right){dx} \\ $$$${A}+{C}=\mathrm{0} \\ $$$$\frac{{A}}{{a}}+{B}+{C}=\mathrm{0} \\ $$$$\frac{{B}}{{a}}+{C}=\mathrm{1} \\ $$$$\Rightarrow\:\frac{{A}}{{a}^{\mathrm{2}} }+\mathrm{1}+{A}−\frac{{A}}{{a}}=\mathrm{0} \\ $$$${A}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{{a}}−\mathrm{1}−\frac{\mathrm{1}}{{a}^{\mathrm{2}} }}=\frac{{a}^{\mathrm{2}} }{{a}−{a}^{\mathrm{2}} −\mathrm{1}} \\ $$$${C}=−{A}\:;\:{B}={a}\left(\mathrm{1}+{A}\right) \\ $$$$\:\:\:\:{B}=\frac{{a}\left({a}−\mathrm{1}\right)}{{a}\left(\mathrm{1}−{a}\right)−\mathrm{1}} \\ $$$$\int\frac{{dx}}{{x}+\frac{\mathrm{1}}{{a}}}=\mathrm{ln}\:\left({x}+\frac{\mathrm{1}}{{a}}\right)+{c}_{\mathrm{1}} \\ $$$$\int\frac{\left(\mathrm{2}{x}+\mathrm{1}\right){dx}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}=\mathrm{ln}\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)+{c}_{\mathrm{2}} \\ $$$$\int\frac{{dx}}{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{c}_{\mathrm{3}} \\ $$

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