Question Number 120562 by peter frank last updated on 01/Nov/20
Answered by Dwaipayan Shikari last updated on 01/Nov/20
$$\int\mathrm{1}−\frac{{bx}+\mathrm{12}}{{x}^{\mathrm{2}} +{bx}+\mathrm{12}}{dx} \\ $$$$={x}−{b}\int\frac{{x}+\frac{\mathrm{12}}{{b}}}{{x}^{\mathrm{2}} +{bx}+\mathrm{12}}{dx} \\ $$$$={x}−\frac{{b}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+{b}}{{x}^{\mathrm{2}} +{bx}+\mathrm{12}}{dx}+\frac{\frac{\mathrm{24}}{{b}}−{b}}{{x}^{\mathrm{2}} +{bx}+\mathrm{12}}{dx} \\ $$$$={x}−\frac{{b}}{\mathrm{2}}{log}\left({x}^{\mathrm{2}} +{bx}+\mathrm{12}\right)−\frac{\mathrm{24}−{b}^{\mathrm{2}} }{\mathrm{2}}\int\frac{\mathrm{1}}{\left({x}+\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{12}−\frac{{b}^{\mathrm{2}} }{\mathrm{4}}}\right)^{\mathrm{2}} }{dx} \\ $$$$={x}−\frac{{b}}{\mathrm{2}}{log}\left({x}^{\mathrm{2}} +{bx}+\mathrm{12}\right)−\frac{\mathrm{24}−{b}^{\mathrm{2}} }{\mathrm{2}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{12}−\frac{{b}^{\mathrm{2}} }{\mathrm{4}}}}{tan}^{−\mathrm{1}} \frac{\mathrm{2}{x}+{b}}{\:\sqrt{\mathrm{48}−{b}^{\mathrm{2}} }} \\ $$$$={x}−\frac{{b}}{\mathrm{2}}{log}\left({x}^{\mathrm{2}} +{bx}+\mathrm{12}\right)−\frac{\mathrm{24}−{b}^{\mathrm{2}} }{\:\sqrt{\mathrm{48}−{b}^{\mathrm{2}} }}{tan}^{−\mathrm{1}} \frac{\mathrm{2}{x}+{b}}{\:\sqrt{\mathrm{48}−{b}^{\mathrm{2}} }}\:+{C} \\ $$
Commented by peter frank last updated on 01/Nov/20
$$\mathrm{thank}\:\mathrm{you} \\ $$