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x-1-x-2-2x-3-2-3-dx-




Question Number 12463 by tawa last updated on 23/Apr/17
∫  ((x + 1)/((x^2  + 2x + 3)^(2/3) ))  dx
$$\int\:\:\frac{\mathrm{x}\:+\:\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{2x}\:+\:\mathrm{3}\right)^{\mathrm{2}/\mathrm{3}} }\:\:\mathrm{dx} \\ $$
Answered by ridwan balatif last updated on 23/Apr/17
∫((x+1)/( (((x^2 +2x+3)^2 ))^(1/3) ))dx  let: x^2 +2x+3=u  (du/dx)=2x+2→dx=(du/(2(x+1)))  ∫((x+1)/( (u^2 )^(1/3) ))×(1/(2(x+1)))du  (1/2)∫u^(−(2/3)) du  (1/2)×(1/3)×u^(1/3) +C  (1/6)(x^2 +2x+3)^(1/3) +C
$$\int\frac{\mathrm{x}+\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{3}\right)^{\mathrm{2}} }}\mathrm{dx} \\ $$$$\mathrm{let}:\:\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{3}=\mathrm{u} \\ $$$$\frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{2x}+\mathrm{2}\rightarrow\mathrm{dx}=\frac{\mathrm{du}}{\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)} \\ $$$$\int\frac{\mathrm{x}+\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{u}^{\mathrm{2}} }}×\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)}\mathrm{du} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{u}^{−\frac{\mathrm{2}}{\mathrm{3}}} \mathrm{du} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{C} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{3}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{C} \\ $$
Commented by tawa last updated on 23/Apr/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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