Question Number 152513 by ZiYangLee last updated on 29/Aug/21
$$\int\:\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{8}}\:{dx}\:=? \\ $$
Commented by alisiao last updated on 29/Aug/21
$$ \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\frac{{dx}}{\left({x}+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{55}}{\mathrm{16}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{{dt}}{{t}^{\mathrm{2}} +\frac{\mathrm{55}}{\mathrm{16}}}\: \\ $$$$ \\ $$$$=\:\frac{\mathrm{8}}{\mathrm{55}}\:\int\:\frac{{dt}}{\mathrm{1}+\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{55}}}\:{t}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{32}}{\mathrm{55}\sqrt{\mathrm{55}}}\:{tan}^{−\mathrm{1}} \left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{55}}}{t}\right)+{C} \\ $$$$ \\ $$$$=\:\frac{\mathrm{32}}{\mathrm{55}\sqrt{\mathrm{55}}}\:{tan}^{−\mathrm{1}} \:\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{55}}}\:\left({x}+\frac{\mathrm{3}}{\mathrm{4}}\right)\right)+{C} \\ $$$$ \\ $$
Answered by puissant last updated on 30/Aug/21
![=(1/2)∫(1/(x^2 +(3/2)+4))dx =(1/2)∫(1/((x+(3/4))^2 −(9/(16))+((64)/(16))))dx =(1/2)∫(1/((x+(3/4))^2 +((55)/(16))))dx = (1/2)∫(1/(((55)/(16))[((4/( (√(55))))(x+(3/4)))^2 +1]))dx =(8/(55))∫(1/(((4/( (√(55))))(x+(3/4)))^2 +1))dx u=(4/( (√(55))))(x+(3/4)) → dx=((√(55))/4)du ⇒ I=(8/(55))×((√(55))/4)∫(du/(u^2 +1)) ⇒ I=(2/( (√(55))))arctan(u)+C ∴∵ Q=(2/( (√(55))))arctan((4/( (√(55))))(x+(3/4)))+C..](https://www.tinkutara.com/question/Q152591.png)
$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{4}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\left({x}+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{16}}+\frac{\mathrm{64}}{\mathrm{16}}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\left({x}+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{55}}{\mathrm{16}}}{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\frac{\mathrm{55}}{\mathrm{16}}\left[\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{55}}}\left({x}+\frac{\mathrm{3}}{\mathrm{4}}\right)\right)^{\mathrm{2}} +\mathrm{1}\right]}{dx} \\ $$$$=\frac{\mathrm{8}}{\mathrm{55}}\int\frac{\mathrm{1}}{\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{55}}}\left({x}+\frac{\mathrm{3}}{\mathrm{4}}\right)\right)^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$${u}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{55}}}\left({x}+\frac{\mathrm{3}}{\mathrm{4}}\right)\:\rightarrow\:{dx}=\frac{\sqrt{\mathrm{55}}}{\mathrm{4}}{du} \\ $$$$\Rightarrow\:{I}=\frac{\mathrm{8}}{\mathrm{55}}×\frac{\sqrt{\mathrm{55}}}{\mathrm{4}}\int\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow\:{I}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{55}}}{arctan}\left({u}\right)+{C} \\ $$$$ \\ $$$$\therefore\because\:{Q}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{55}}}{arctan}\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{55}}}\left({x}+\frac{\mathrm{3}}{\mathrm{4}}\right)\right)+{C}.. \\ $$