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Question Number 35683 by prof Abdo imad last updated on 22/May/18
find ∫ x^2 ln(x^6 −1)dx
$${find}\:\int\:\:{x}^{\mathrm{2}} {ln}\left({x}^{\mathrm{6}} −\mathrm{1}\right){dx} \\ $$
Commented by prof Abdo imad last updated on 30/May/18
let put I = ∫ x^2 ln(x^6 −1)dx I = ∫ x^2 ln{ (x^3 +1)(x^3 −1)}dx = ∫ x^2 ln(x^3 +1)dx + ∫ x^2 ln(x^3 −1)dx= I_1 +I_2 by parts I_1 = (x^3 /3)ln(x^3 +1) −∫ (x^3 /3) ((3x^2 )/(x^3 +1))dx = (x^3 /3)ln(x^3 +1) −∫ (x^5 /(x^3 +1))dx but ∫ (x^5 /(x^3 +1))dx = ∫ ((x^2 (x^3 +1)−x^2 )/(x^3 +1))dx =(x^3 /3) − ∫ (x^2 /(x^3 +1))dx let decompose F(x)= (x^2 /(x^3 +1)) = (x^2 /((x+1)(x^2 −x+1))) = (a/(x+1)) +((bx+c)/(x^2 −x +1)) ....be continued...
$${let}\:{put}\:{I}\:=\:\int\:{x}^{\mathrm{2}} {ln}\left({x}^{\mathrm{6}} −\mathrm{1}\right){dx} \\ $$$${I}\:=\:\int\:{x}^{\mathrm{2}} {ln}\left\{\:\left({x}^{\mathrm{3}} +\mathrm{1}\right)\left({x}^{\mathrm{3}} −\mathrm{1}\right)\right\}{dx} \\ $$$$=\:\int\:{x}^{\mathrm{2}} {ln}\left({x}^{\mathrm{3}} +\mathrm{1}\right){dx}\:+\:\int\:{x}^{\mathrm{2}} {ln}\left({x}^{\mathrm{3}} \:−\mathrm{1}\right){dx}=\:{I}_{\mathrm{1}} \:+{I}_{\mathrm{2}} \\ $$$$\:{by}\:{parts}\: \\ $$$${I}_{\mathrm{1}} =\:\frac{{x}^{\mathrm{3}} }{\mathrm{3}}{ln}\left({x}^{\mathrm{3}} +\mathrm{1}\right)\:−\int\:\:\:\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:\frac{\mathrm{3}{x}^{\mathrm{2}} }{{x}^{\mathrm{3}} \:+\mathrm{1}}{dx} \\ $$$$=\:\frac{{x}^{\mathrm{3}} }{\mathrm{3}}{ln}\left({x}^{\mathrm{3}} +\mathrm{1}\right)\:\:−\int\:\:\frac{{x}^{\mathrm{5}} }{{x}^{\mathrm{3}} \:+\mathrm{1}}{dx}\:{but} \\ $$$$\int\:\:\:\frac{{x}^{\mathrm{5}} }{{x}^{\mathrm{3}} +\mathrm{1}}{dx}\:=\:\int\:\frac{{x}^{\mathrm{2}} \left({x}^{\mathrm{3}} +\mathrm{1}\right)−{x}^{\mathrm{2}} }{{x}^{\mathrm{3}} \:+\mathrm{1}}{dx} \\ $$$$=\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:−\:\int\:\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{3}} +\mathrm{1}}{dx}\:{let}\:{decompose} \\ $$$${F}\left({x}\right)=\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{3}} \:+\mathrm{1}}\:=\:\frac{{x}^{\mathrm{2}} }{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:−{x}+\mathrm{1}\right)}\: \\ $$$$=\:\frac{{a}}{{x}+\mathrm{1}}\:+\frac{{bx}+{c}}{{x}^{\mathrm{2}} \:−{x}\:+\mathrm{1}}\:\:\:….{be}\:{continued}… \\ $$$$ \\ $$
Answered by sma3l2996 last updated on 31/May/18
A=∫x^2 ln(x^6 −1)dx let t=x^3 ⇒dt=3x^2 dx A=(1/3)∫ln(t^2 −1)dt by parts A=(1/3)×t×ln(t^2 −1)−(2/3)∫(t^2 /(t^2 −1))dt+c =((x^3 ln(x^6 −1))/3)−(2/3)∫(1+(1/((t−1)(t+1))))dt+c =((x^3 ln(x^6 −1))/3)−(2/3)x^3 −(1/3)∫((1/(t−1))−(1/(t+1)))dt+c_1 =((x^3 ln(x^6 −1))/3)−((2x^3 )/3)+(1/3)ln∣((t+1)/(t−1))∣+C A=((x^3 ln(x^6 −1))/3)−((2x^3 )/3)+(1/3)ln∣((x^3 +1)/(x^3 −1))∣+C
$${A}=\int{x}^{\mathrm{2}} {ln}\left({x}^{\mathrm{6}} −\mathrm{1}\right){dx} \\ $$$${let}\:{t}={x}^{\mathrm{3}} \Rightarrow{dt}=\mathrm{3}{x}^{\mathrm{2}} {dx} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{3}}\int{ln}\left({t}^{\mathrm{2}} −\mathrm{1}\right){dt} \\ $$$${by}\:{parts} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{3}}×{t}×{ln}\left({t}^{\mathrm{2}} −\mathrm{1}\right)−\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} −\mathrm{1}}{dt}+{c} \\ $$$$=\frac{{x}^{\mathrm{3}} {ln}\left({x}^{\mathrm{6}} −\mathrm{1}\right)}{\mathrm{3}}−\frac{\mathrm{2}}{\mathrm{3}}\int\left(\mathrm{1}+\frac{\mathrm{1}}{\left({t}−\mathrm{1}\right)\left({t}+\mathrm{1}\right)}\right){dt}+{c} \\ $$$$=\frac{{x}^{\mathrm{3}} {ln}\left({x}^{\mathrm{6}} −\mathrm{1}\right)}{\mathrm{3}}−\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{3}}\int\left(\frac{\mathrm{1}}{{t}−\mathrm{1}}−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right){dt}+{c}_{\mathrm{1}} \\ $$$$=\frac{{x}^{\mathrm{3}} {ln}\left({x}^{\mathrm{6}} −\mathrm{1}\right)}{\mathrm{3}}−\frac{\mathrm{2}{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid\frac{{t}+\mathrm{1}}{{t}−\mathrm{1}}\mid+{C} \\ $$$${A}=\frac{{x}^{\mathrm{3}} {ln}\left({x}^{\mathrm{6}} −\mathrm{1}\right)}{\mathrm{3}}−\frac{\mathrm{2}{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid\frac{{x}^{\mathrm{3}} +\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{1}}\mid+{C} \\ $$

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