Question Number 20908 by j.masanja06@gmail.com last updated on 07/Sep/17
$${integrate}\:{with}\:{respect}\:{to}\:{x}\: \\ $$$$\int\left(\frac{\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{8}}\right){dx} \\ $$
Answered by Joel577 last updated on 07/Sep/17
$${I}\:=\:\int\:\frac{\mathrm{2}{x}\:+\:\mathrm{1}}{\left({x}\:+\:\mathrm{2}\right)^{\mathrm{2}} \:+\:\mathrm{4}}\:{dx} \\ $$$$ \\ $$$$\mathrm{Let}\:{u}\:=\:{x}\:+\:\mathrm{2}\:\:\rightarrow\:\:{du}\:=\:{dx} \\ $$$${I}\:=\:\int\:\frac{\mathrm{2}{u}\:−\:\mathrm{3}}{{u}^{\mathrm{2}} \:+\:\mathrm{4}}\:{du} \\ $$$$ \\ $$$$\mathrm{Let}\:{u}\:=\:\mathrm{2tan}\:\theta\:\:\rightarrow\:\:{du}\:=\:\mathrm{2sec}^{\mathrm{2}} \:\theta\:{d}\theta \\ $$$$\:\mathrm{tan}\:\theta\:=\:\frac{{u}}{\mathrm{2}}\:\:\rightarrow\:\:\theta\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{u}}{\mathrm{2}}\right) \\ $$$${I}\:=\:\int\:\frac{\mathrm{4tan}\:\theta\:−\:\mathrm{3}}{\mathrm{4}\left(\mathrm{tan}^{\mathrm{2}} \:\theta\:+\:\mathrm{1}\right)}\:\mathrm{2sec}^{\mathrm{2}} \:\theta\:{d}\theta\: \\ $$$$\:\:\:=\:\int\:\frac{\mathrm{4tan}\:\theta\:−\:\mathrm{3}}{\mathrm{2}}\:{d}\theta\:=\:\int\:\mathrm{2tan}\:\theta\:−\:\frac{\mathrm{3}}{\mathrm{2}}\:{d}\theta \\ $$$$\:\:\:=\:−\mathrm{2ln}\:\mid\mathrm{cos}\:\theta\mid\:−\:\frac{\mathrm{3}}{\mathrm{2}}\theta\:+\:{C} \\ $$$$\:\:\:=\:−\mathrm{2ln}\:\mid\:\frac{\mathrm{2}}{\:\sqrt{{u}^{\mathrm{2}} \:+\:\mathrm{4}}}\:\mid\:−\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{u}}{\mathrm{2}}\right)\:+\:{C} \\ $$$$\:\:\:=\:−\mathrm{2ln}\:\mid\:\frac{\mathrm{2}}{\:\sqrt{\left({x}\:+\:\mathrm{2}\right)^{\mathrm{2}} \:+\:\mathrm{4}}}\:\mid\:−\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}\:+\:\mathrm{2}}{\mathrm{2}}\right)\:+\:{C} \\ $$