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Question Number 120562 by peter frank last updated on 01/Nov/20

Answered by Dwaipayan Shikari last updated on 01/Nov/20

∫1−((bx+12)/(x^2 +bx+12))dx  =x−b∫((x+((12)/b))/(x^2 +bx+12))dx  =x−(b/2)∫((2x+b)/(x^2 +bx+12))dx+((((24)/b)−b)/(x^2 +bx+12))dx  =x−(b/2)log(x^2 +bx+12)−((24−b^2 )/2)∫(1/((x+(b/2))^2 +((√(12−(b^2 /4))))^2 ))dx  =x−(b/2)log(x^2 +bx+12)−((24−b^2 )/2).(1/( (√(12−(b^2 /4)))))tan^(−1) ((2x+b)/( (√(48−b^2 ))))  =x−(b/2)log(x^2 +bx+12)−((24−b^2 )/( (√(48−b^2 ))))tan^(−1) ((2x+b)/( (√(48−b^2 )))) +C

$$\int\mathrm{1}−\frac{{bx}+\mathrm{12}}{{x}^{\mathrm{2}} +{bx}+\mathrm{12}}{dx} \\ $$$$={x}−{b}\int\frac{{x}+\frac{\mathrm{12}}{{b}}}{{x}^{\mathrm{2}} +{bx}+\mathrm{12}}{dx} \\ $$$$={x}−\frac{{b}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+{b}}{{x}^{\mathrm{2}} +{bx}+\mathrm{12}}{dx}+\frac{\frac{\mathrm{24}}{{b}}−{b}}{{x}^{\mathrm{2}} +{bx}+\mathrm{12}}{dx} \\ $$$$={x}−\frac{{b}}{\mathrm{2}}{log}\left({x}^{\mathrm{2}} +{bx}+\mathrm{12}\right)−\frac{\mathrm{24}−{b}^{\mathrm{2}} }{\mathrm{2}}\int\frac{\mathrm{1}}{\left({x}+\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{12}−\frac{{b}^{\mathrm{2}} }{\mathrm{4}}}\right)^{\mathrm{2}} }{dx} \\ $$$$={x}−\frac{{b}}{\mathrm{2}}{log}\left({x}^{\mathrm{2}} +{bx}+\mathrm{12}\right)−\frac{\mathrm{24}−{b}^{\mathrm{2}} }{\mathrm{2}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{12}−\frac{{b}^{\mathrm{2}} }{\mathrm{4}}}}{tan}^{−\mathrm{1}} \frac{\mathrm{2}{x}+{b}}{\:\sqrt{\mathrm{48}−{b}^{\mathrm{2}} }} \\ $$$$={x}−\frac{{b}}{\mathrm{2}}{log}\left({x}^{\mathrm{2}} +{bx}+\mathrm{12}\right)−\frac{\mathrm{24}−{b}^{\mathrm{2}} }{\:\sqrt{\mathrm{48}−{b}^{\mathrm{2}} }}{tan}^{−\mathrm{1}} \frac{\mathrm{2}{x}+{b}}{\:\sqrt{\mathrm{48}−{b}^{\mathrm{2}} }}\:+{C} \\ $$

Commented by peter frank last updated on 01/Nov/20

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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