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Question Number 20908 by j.masanja06@gmail.com last updated on 07/Sep/17

integrate with respect to x ∫(((2x+1)/(x^2 +4x+8)))dx

$${integrate}\:{with}\:{respect}\:{to}\:{x}\: \\ $$$$\int\left(\frac{\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{8}}\right){dx} \\ $$

Answered by Joel577 last updated on 07/Sep/17

I = ∫ ((2x + 1)/((x + 2)^2 + 4)) dx Let u = x + 2 → du = dx I = ∫ ((2u − 3)/(u^2 + 4)) du Let u = 2tan θ → du = 2sec^2 θ dθ tan θ = (u/2) → θ = tan^(−1) ((u/2)) I = ∫ ((4tan θ − 3)/(4(tan^2 θ + 1))) 2sec^2 θ dθ = ∫ ((4tan θ − 3)/2) dθ = ∫ 2tan θ − (3/2) dθ = −2ln ∣cos θ∣ − (3/2)θ + C = −2ln ∣ (2/(√(u^2 + 4))) ∣ − (3/2)tan^(−1) ((u/2)) + C = −2ln ∣ (2/(√((x + 2)^2 + 4))) ∣ − (3/2)tan^(−1) (((x + 2)/2)) + C

$${I}\:=\:\int\:\frac{\mathrm{2}{x}\:+\:\mathrm{1}}{\left({x}\:+\:\mathrm{2}\right)^{\mathrm{2}} \:+\:\mathrm{4}}\:{dx} \\ $$$$ \\ $$$$\mathrm{Let}\:{u}\:=\:{x}\:+\:\mathrm{2}\:\:\rightarrow\:\:{du}\:=\:{dx} \\ $$$${I}\:=\:\int\:\frac{\mathrm{2}{u}\:−\:\mathrm{3}}{{u}^{\mathrm{2}} \:+\:\mathrm{4}}\:{du} \\ $$$$ \\ $$$$\mathrm{Let}\:{u}\:=\:\mathrm{2tan}\:\theta\:\:\rightarrow\:\:{du}\:=\:\mathrm{2sec}^{\mathrm{2}} \:\theta\:{d}\theta \\ $$$$\:\mathrm{tan}\:\theta\:=\:\frac{{u}}{\mathrm{2}}\:\:\rightarrow\:\:\theta\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{u}}{\mathrm{2}}\right) \\ $$$${I}\:=\:\int\:\frac{\mathrm{4tan}\:\theta\:−\:\mathrm{3}}{\mathrm{4}\left(\mathrm{tan}^{\mathrm{2}} \:\theta\:+\:\mathrm{1}\right)}\:\mathrm{2sec}^{\mathrm{2}} \:\theta\:{d}\theta\: \\ $$$$\:\:\:=\:\int\:\frac{\mathrm{4tan}\:\theta\:−\:\mathrm{3}}{\mathrm{2}}\:{d}\theta\:=\:\int\:\mathrm{2tan}\:\theta\:−\:\frac{\mathrm{3}}{\mathrm{2}}\:{d}\theta \\ $$$$\:\:\:=\:−\mathrm{2ln}\:\mid\mathrm{cos}\:\theta\mid\:−\:\frac{\mathrm{3}}{\mathrm{2}}\theta\:+\:{C} \\ $$$$\:\:\:=\:−\mathrm{2ln}\:\mid\:\frac{\mathrm{2}}{\sqrt{{u}^{\mathrm{2}} \:+\:\mathrm{4}}}\:\mid\:−\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{u}}{\mathrm{2}}\right)\:+\:{C} \\ $$$$\:\:\:=\:−\mathrm{2ln}\:\mid\:\frac{\mathrm{2}}{\sqrt{\left({x}\:+\:\mathrm{2}\right)^{\mathrm{2}} \:+\:\mathrm{4}}}\:\mid\:−\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}\:+\:\mathrm{2}}{\mathrm{2}}\right)\:+\:{C} \\ $$

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