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Question Number 33689 by NECx last updated on 22/Apr/18
∫(x/(x^3 +1))dx
$$\int\frac{{x}}{{x}^{\mathrm{3}} +\mathrm{1}}{dx} \\ $$
Commented by mondodotto@gmail.com last updated on 22/Apr/18
let u=x^3 +1 (du/dx)=3x^2 ⇒dx=(du/(3x^2 )) x=((u−1))^(1/3) ∴ ∫(x/(x^3 +1))dx⇒ ∫(x/u)(du/(3x^2 ))⇒∫(du/(3ux)) ⇒(1/3)∫(du/(u(((u−1))^(1/3) ))) continue...
$$\boldsymbol{\mathrm{let}}\:\boldsymbol{\mathrm{u}}=\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{1} \\ $$$$\frac{\boldsymbol{\mathrm{du}}}{\boldsymbol{\mathrm{dx}}}=\mathrm{3}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \Rightarrow\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\mathrm{du}}}{\mathrm{3}\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \\ $$$$\boldsymbol{\mathrm{x}}=\sqrt[{\mathrm{3}}]{\boldsymbol{\mathrm{u}}−\mathrm{1}} \\ $$$$\therefore\:\int\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{1}}\boldsymbol{\mathrm{dx}}\Rightarrow\:\int\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{u}}}\frac{\boldsymbol{\mathrm{du}}}{\mathrm{3}\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\Rightarrow\int\frac{\boldsymbol{\mathrm{du}}}{\mathrm{3}\boldsymbol{\mathrm{ux}}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\boldsymbol{\mathrm{du}}}{\boldsymbol{\mathrm{u}}\left(\sqrt[{\mathrm{3}}]{\boldsymbol{\mathrm{u}}−\mathrm{1}}\right)} \\ $$$$\boldsymbol{\mathrm{continue}}… \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Apr/18
∫(x+1−1)/x^3 +1 dx ∫dx/(x^2 −x+1) −∫dx/(x^3 +1) I1=∫dx/x^2 −2.x.1/2+1/4 +1−1/4 =∫dx/(x−1/2)^2 +((√)3/2)^2 use formula∫dx/(x^2 +a^2 ) I2∫dx/(x^3 +1) 1/(x^3 +1)=A/(x+1) +(Bx+C)/(x^2 −x+1) 1 =A(x^2 −x+1)+(x+1)(Bx+C) x^2 (A+B) +x(−A+B+C) +(A+C) so A+B=0 −A+B+C=0 A+C=1 solve and proceed if any problem pls comment
$$\int\left({x}+\mathrm{1}−\mathrm{1}\right)/{x}^{\mathrm{3}} +\mathrm{1}\:\:{dx} \\ $$$$\int{dx}/\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:\:−\int{dx}/\left({x}^{\mathrm{3}} +\mathrm{1}\right) \\ $$$${I}\mathrm{1}=\int{dx}/{x}^{\mathrm{2}} −\mathrm{2}.{x}.\mathrm{1}/\mathrm{2}+\mathrm{1}/\mathrm{4}\:+\mathrm{1}−\mathrm{1}/\mathrm{4} \\ $$$$=\int{dx}/\left({x}−\mathrm{1}/\mathrm{2}\right)^{\mathrm{2}} +\left(\sqrt{}\mathrm{3}/\mathrm{2}\right)^{\mathrm{2}} \: \\ $$$${use}\:{formula}\int{dx}/\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right) \\ $$$$\mathrm{I2}\int{dx}/\left({x}^{\mathrm{3}} +\mathrm{1}\right) \\ $$$$\mathrm{1}/\left({x}^{\mathrm{3}} +\mathrm{1}\right)={A}/\left({x}+\mathrm{1}\right)\:+\left({Bx}+{C}\right)/\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:={A}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)+\left({x}+\mathrm{1}\right)\left({Bx}+{C}\right) \\ $$$${x}^{\mathrm{2}} \left({A}+{B}\right)\:+{x}\left(−{A}+{B}+{C}\right)\:+\left({A}+{C}\right) \\ $$$${so}\:{A}+{B}=\mathrm{0} \\ $$$$−{A}+{B}+{C}=\mathrm{0} \\ $$$${A}+{C}=\mathrm{1} \\ $$$${solve}\:{and}\:{proceed}\:{if}\:{any}\:{problem}\:{pls}\:{comment} \\ $$$$ \\ $$$$ \\ $$
Answered by math1967 last updated on 22/Apr/18
(1/3)∫((3x)/((x+1)(x^2 −x+1)))dx (1/3)∫(((x+1)^2 −(x^2 −x+1))/((x+1)(x^2 −x+1)))dx (1/3)∫((x+1)/(x^2 −x+1))dx −(1/3)∫(dx/((x+1))) (1/6)∫((2x+2)/(x^2 −x+1))dx −(1/3)ln[x+1] (1/6)∫((2x−1)/(x^2 −x+1))dx +(1/6)∫(3/(x^2 −x+1))dx −(1/3)ln[x+1] (1/6)∫((d(x^2 −x+1))/(x^2 −x+1)) +(1/2)∫(dx/((x−(1/2))^2 +(((√(3 ))/2))^2 )) −(1/3)ln[x+1] (1/6)ln[x^2 −x+1] +(1/2)×(2/( (√3)))tan^(−1) (((x−(1/2))/((√3)/2))) −(1/3)ln[x+1] +c
$$\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{3}{x}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}\:−\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{\left({x}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\mathrm{2}{x}+\mathrm{2}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}\:−\frac{\mathrm{1}}{\mathrm{3}}{ln}\left[{x}+\mathrm{1}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}\:+\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\mathrm{3}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}\:−\frac{\mathrm{1}}{\mathrm{3}}{ln}\left[{x}+\mathrm{1}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\int\frac{{d}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}\:\:}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{3}}{ln}\left[{x}+\mathrm{1}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}{ln}\left[{x}^{\mathrm{2}} −{x}+\mathrm{1}\right]\:+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}−\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{3}}{ln}\left[{x}+\mathrm{1}\right]\:+{c} \\ $$$$\:\: \\ $$
Commented by Joel578 last updated on 22/Apr/18
(1/6)ln (x^2 − x + 1) − (1/3)ln ∣x + 1∣ + (1/( (√3)))tan^(−1) (((2x −1)/( (√3)))) + C
$$\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\left({x}^{\mathrm{2}} \:−\:{x}\:+\:\mathrm{1}\right)\:−\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\mid{x}\:+\:\mathrm{1}\mid\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}\:−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:+\:{C} \\ $$

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