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Question Number 36430 by prof Abdo imad last updated on 02/Jun/18
find ∫ ((ln(x+x^2 ))/x^2 )dx
$${find}\:\:\int\:\:\:\frac{{ln}\left({x}+{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx}\: \\ $$
Commented by abdo mathsup last updated on 03/Jun/18
let integrate by parts u^′ =(1/x^2 ) and v=ln(x+x^2 ) I = −(1/x)ln(x+x^2 ) +∫ (1/x) ((2x+1)/(x+x^2 ))dx =−((ln(x+x^2 ))/x) + ∫ ((2x+1)/(x^2 (x+1)))dx but ∫ ((2x+1)/(x^2 (x+1)))dx = ∫ ((2(x+1) −1)/(x^2 (x+1)))dx = 2 ∫ (dx/x^2 ) −∫ (dx/(x^2 (x+1))) =((−2)/x) −∫ (dx/(x^2 (x+1))) F(x)= (1/(x^2 (x+1))) =(a/x) +(b/x^2 ) +(c/(x+1)) b=lim_(x→0) x^2 F(x) =1 c=lim_(x→−1) (x+1)F(x)= 1 ⇒ F(x) = (1/x) +(b/x^2 ) +(1/(x+1)) F(1) = (1/2) = 1 +b + (1/2) ⇒b=−1 ⇒ F(x) = (1/x) −(1/x^2 ) + (1/(x+1)) ⇒ ∫ F(x)dx = ln∣x∣ +(1/x) −ln∣x+1∣ +c ⇒ I =−((ln(x^2 +x))/x) −(2/x) −ln∣x∣ −(1/x) +ln∣x+1∣ +c⇒ I = −((ln(x^2 +x))/x) −(3/x) −ln∣x∣ +ln∣x+1∣ +c .
$${let}\:{integrate}\:{by}\:{parts}\:\:{u}^{'} \:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:{and}\:{v}={ln}\left({x}+{x}^{\mathrm{2}} \right) \\ $$$${I}\:=\:−\frac{\mathrm{1}}{{x}}{ln}\left({x}+{x}^{\mathrm{2}} \right)\:\:+\int\:\:\:\frac{\mathrm{1}}{{x}}\:\frac{\mathrm{2}{x}+\mathrm{1}}{{x}+{x}^{\mathrm{2}} }{dx} \\ $$$$=−\frac{{ln}\left({x}+{x}^{\mathrm{2}} \right)}{{x}}\:\:+\:\int\:\:\:\frac{\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)}{dx}\:{but} \\ $$$$\int\:\:\frac{\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)}{dx}\:=\:\int\:\:\frac{\mathrm{2}\left({x}+\mathrm{1}\right)\:−\mathrm{1}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)}{dx} \\ $$$$=\:\mathrm{2}\:\int\:\frac{{dx}}{{x}^{\mathrm{2}} }\:\:−\int\:\:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)}\:=\frac{−\mathrm{2}}{{x}}\:−\int\:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)} \\ $$$${F}\left({x}\right)=\:\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)}\:=\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }\:\:+\frac{{c}}{{x}+\mathrm{1}} \\ $$$${b}={lim}_{{x}\rightarrow\mathrm{0}} {x}^{\mathrm{2}} {F}\left({x}\right)\:=\mathrm{1}\: \\ $$$${c}={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)=\:\mathrm{1}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\:\:\frac{\mathrm{1}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }\:\:+\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$$${F}\left(\mathrm{1}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:\mathrm{1}\:\:+{b}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{b}=−\mathrm{1}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\:\frac{\mathrm{1}}{{x}}\:\:−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{x}+\mathrm{1}}\:\Rightarrow \\ $$$$\int\:{F}\left({x}\right){dx}\:=\:{ln}\mid{x}\mid\:\:+\frac{\mathrm{1}}{{x}}\:−{ln}\mid{x}+\mathrm{1}\mid\:+{c}\:\Rightarrow \\ $$$${I}\:=−\frac{{ln}\left({x}^{\mathrm{2}} \:+{x}\right)}{{x}}\:\:−\frac{\mathrm{2}}{{x}}\:−{ln}\mid{x}\mid\:−\frac{\mathrm{1}}{{x}}\:+{ln}\mid{x}+\mathrm{1}\mid\:+{c}\Rightarrow \\ $$$${I}\:=\:−\frac{{ln}\left({x}^{\mathrm{2}} \:+{x}\right)}{{x}}\:−\frac{\mathrm{3}}{{x}}\:\:−{ln}\mid{x}\mid\:+{ln}\mid{x}+\mathrm{1}\mid\:+{c}\:. \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18
ln(x+x^2 )×((−1)/x)−∫(1/(x+x^2 ))×(1+2x)×((−1)/x)dx ((−ln(x+x^2 ))/x)+∫((1+2x)/(x^2 (1+x)))dx I_2 =∫((1+2x)/(x^2 (1+x)))dx =∫((1+x+x)/(x^2 (1+x)))dx =∫(dx/x^2 )+∫(dx/(x(1+x))) =∫x^(−2) dx+∫((1+x−x)/(x(1+x)))dx =((−1)/x)+∫(dx/x)−∫(dx/(1+x)) =((−1)/x)+ln((x/(1+x))) =−((ln(x+x^2 ))/x)−(1/x)+ln(x/(1+x))
$${ln}\left({x}+{x}^{\mathrm{2}} \right)×\frac{−\mathrm{1}}{{x}}−\int\frac{\mathrm{1}}{{x}+{x}^{\mathrm{2}} }×\left(\mathrm{1}+\mathrm{2}{x}\right)×\frac{−\mathrm{1}}{{x}}{dx} \\ $$$$\frac{−{ln}\left({x}+{x}^{\mathrm{2}} \right)}{{x}}+\int\frac{\mathrm{1}+\mathrm{2}{x}}{{x}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)}{dx} \\ $$$${I}_{\mathrm{2}} =\int\frac{\mathrm{1}+\mathrm{2}{x}}{{x}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)}{dx} \\ $$$$=\int\frac{\mathrm{1}+{x}+{x}}{{x}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)}{dx} \\ $$$$=\int\frac{{dx}}{{x}^{\mathrm{2}} }+\int\frac{{dx}}{{x}\left(\mathrm{1}+{x}\right)} \\ $$$$=\int{x}^{−\mathrm{2}} {dx}+\int\frac{\mathrm{1}+{x}−{x}}{{x}\left(\mathrm{1}+{x}\right)}{dx} \\ $$$$=\frac{−\mathrm{1}}{{x}}+\int\frac{{dx}}{{x}}−\int\frac{{dx}}{\mathrm{1}+{x}} \\ $$$$=\frac{−\mathrm{1}}{{x}}+{ln}\left(\frac{{x}}{\mathrm{1}+{x}}\right) \\ $$$$=−\frac{{ln}\left({x}+{x}^{\mathrm{2}} \right)}{{x}}−\frac{\mathrm{1}}{{x}}+{ln}\frac{{x}}{\mathrm{1}+{x}} \\ $$$$ \\ $$

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