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