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Question Number 31787 by neel1974 last updated on 14/Mar/18

∫((4x−3)/(x^2 +3x+8))dx

$$\int\frac{\mathrm{4}{x}−\mathrm{3}}{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{8}}{dx} \\ $$

Answered by sma3l2996 last updated on 14/Mar/18

A=∫((4x−3)/(x^2 +3x+8))dx=2∫((2x−3/2)/(x^2 +3x+8))dx =2∫(((2x+3)/(x^2 +3x+8))−((3+3/2)/(x^2 +3x+8)))dx =2ln(x^2 +3x+8)−9∫(dx/(x^2 +3x+8))+c ∫(dx/(x^2 +3x+8))=∫(dx/(x^2 +2×(3/2)×x+((3/2))^2 −(9/4)+8)) =∫(dx/((x+(3/2))^2 +((23)/4)))=(4/(23))∫(dx/((((2x+3)/(√(23))))^2 +1)) let t=((2x+3)/(√(23)))⇒dt=(2/(√(23)))dx ∫(dx/(x^2 +3x+8))=((2(√(23)))/(23))∫(dt/(t^2 +1))=((2(√(23)))/(23))tan^(−1) (((2x+3)/(√(23))))+k so A=2ln(x^2 +3x+8)−((18(√(23)))/(23))tan^(−1) (((2x+3)/(√(23))))+C

$${A}=\int\frac{\mathrm{4}{x}−\mathrm{3}}{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{8}}{dx}=\mathrm{2}\int\frac{\mathrm{2}{x}−\mathrm{3}/\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{8}}{dx} \\ $$$$=\mathrm{2}\int\left(\frac{\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{8}}−\frac{\mathrm{3}+\mathrm{3}/\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{8}}\right){dx} \\ $$$$=\mathrm{2}{ln}\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{8}\right)−\mathrm{9}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{8}}+{c} \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{8}}=\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{2}×\frac{\mathrm{3}}{\mathrm{2}}×{x}+\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}}+\mathrm{8}} \\ $$$$=\int\frac{{dx}}{\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{23}}{\mathrm{4}}}=\frac{\mathrm{4}}{\mathrm{23}}\int\frac{{dx}}{\left(\frac{\mathrm{2}{x}+\mathrm{3}}{\sqrt{\mathrm{23}}}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$${let}\:\:{t}=\frac{\mathrm{2}{x}+\mathrm{3}}{\sqrt{\mathrm{23}}}\Rightarrow{dt}=\frac{\mathrm{2}}{\sqrt{\mathrm{23}}}{dx} \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{8}}=\frac{\mathrm{2}\sqrt{\mathrm{23}}}{\mathrm{23}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{2}\sqrt{\mathrm{23}}}{\mathrm{23}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}+\mathrm{3}}{\sqrt{\mathrm{23}}}\right)+{k} \\ $$$${so}\: \\ $$$${A}=\mathrm{2}{ln}\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{8}\right)−\frac{\mathrm{18}\sqrt{\mathrm{23}}}{\mathrm{23}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}+\mathrm{3}}{\sqrt{\mathrm{23}}}\right)+{C} \\ $$

Answered by ajfour last updated on 14/Mar/18

I=2∫((2x+3)/(x^2 +3x+8))dx−9∫(dx/((x+(3/2))^2 +(((√(23))/2))^2 )) I=2ln ∣x^2 +3x+8∣−((18)/(√(23)))tan^(−1) (((2x+3)/(√(23))))+c .

$${I}=\mathrm{2}\int\frac{\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{8}}{dx}−\mathrm{9}\int\frac{{dx}}{\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{23}}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${I}=\mathrm{2ln}\:\mid{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{8}\mid−\frac{\mathrm{18}}{\sqrt{\mathrm{23}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}+\mathrm{3}}{\sqrt{\mathrm{23}}}\right)+{c}\:. \\ $$

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