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Question Number 171971 by ilhamQ last updated on 22/Jun/22

∫(x/(x^2 +4x+3)) dx=...

$$\int\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}}\:{dx}=... \\ $$

Answered by cortano1 last updated on 22/Jun/22

(x/((x+1)(x+3))) = (a/(x+1)) + (b/(x+3)) a = ((−1)/(−1+3)) =−(1/2) b=((−3)/(−3+1)) = (3/2) ∫ (x/(x^2 +4x+3)) dx = −(1/2) ln ∣x+1∣+(3/2)ln ∣x+3∣ +c

$$\:\:\frac{{x}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right)}\:=\:\frac{{a}}{{x}+\mathrm{1}}\:+\:\frac{{b}}{{x}+\mathrm{3}} \\ $$$$\:{a}\:=\:\frac{−\mathrm{1}}{−\mathrm{1}+\mathrm{3}}\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:{b}=\frac{−\mathrm{3}}{−\mathrm{3}+\mathrm{1}}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\:\int\:\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}}\:{dx}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{ln}\:\mid{x}+\mathrm{1}\mid+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln}\:\mid{x}+\mathrm{3}\mid\:+{c}\: \\ $$

Answered by Mathspace last updated on 22/Jun/22

I=(1/2)∫((2x+4−4)/(x^2 +4x+3))dx =(1/2)∫((2x+4)/(x^2 +4x+3))dx−2∫(dx/(x^2 +4x+3)) x^2 +4x+3=0 →Δ^′ =2^2 −3=1 ⇒ x_1 =((−2+1)/1)=−1 and x_2 =((−2−1)/1)=−3 ⇒ (1/(x^2 +4x+3))=(1/((x+1)(x+3))) =(1/2)((1/(x+1))−(1/(x+3)))⇒ ∫(dx/(x^2 +4x+3))=(1/2)ln∣((x+1)/(x+3))∣ +c_1 ∫((2x+4)/(x^2 +4x+3))dx=ln∣x^2 +4x+3∣+c_2 ⇒I=(1/2)ln∣x^2 +4x+3∣−ln∣((x+1)/(x+3))∣ +C

$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\mathrm{4}−\mathrm{4}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\mathrm{4}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}}{dx}−\mathrm{2}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}} \\ $$$${x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}=\mathrm{0}\:\rightarrow\Delta^{'} =\mathrm{2}^{\mathrm{2}} −\mathrm{3}=\mathrm{1}\:\Rightarrow \\ $$$${x}_{\mathrm{1}} =\frac{−\mathrm{2}+\mathrm{1}}{\mathrm{1}}=−\mathrm{1}\:{and}\:{x}_{\mathrm{2}} =\frac{−\mathrm{2}−\mathrm{1}}{\mathrm{1}}=−\mathrm{3}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}}=\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{{x}+\mathrm{3}}\right)\Rightarrow \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{{x}+\mathrm{1}}{{x}+\mathrm{3}}\mid\:+{c}_{\mathrm{1}} \\ $$$$\int\frac{\mathrm{2}{x}+\mathrm{4}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}}{dx}={ln}\mid{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}\mid+{c}_{\mathrm{2}} \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}\mid−{ln}\mid\frac{{x}+\mathrm{1}}{{x}+\mathrm{3}}\mid\:+{C} \\ $$

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