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




Question Number 129794 by bramlexs22 last updated on 19/Jan/21
  ∫ (x^2 −1)(x+1)^(−2/3)  dx ?
$$\:\:\int\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)^{−\mathrm{2}/\mathrm{3}} \:\mathrm{dx}\:? \\ $$
Answered by EDWIN88 last updated on 19/Jan/21
 ∫ (x−1)(x+1)^(1/3)  dx = (x−1)((3/4)(x+1)^(4/3) )−(3/4)∫(x+1)^(4/3)  dx   = (3/4)(x−1)(x+1)^(4/3) −(9/(28))(x+1)^(7/3)  + C
$$\:\int\:\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} \:\mathrm{dx}\:=\:\left(\mathrm{x}−\mathrm{1}\right)\left(\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{4}/\mathrm{3}} \right)−\frac{\mathrm{3}}{\mathrm{4}}\int\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{4}/\mathrm{3}} \:\mathrm{dx} \\ $$$$\:=\:\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{4}/\mathrm{3}} −\frac{\mathrm{9}}{\mathrm{28}}\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{7}/\mathrm{3}} \:+\:\mathrm{C} \\ $$$$ \\ $$
Answered by Lordose last updated on 19/Jan/21
I = ∫(x+1)(x−1)(x+1)^(−(2/3))  = ∫(x+1)^(1/3) (x−1)dx  I =^(u=x+1) ∫u^(1/3) (u−2)du = ∫u^(4/3) du − 2∫u^(1/3) du   I = ((3u^(7/3) )/7) − ((3u^(4/3) )/2) + C
$$\mathrm{I}\:=\:\int\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)^{−\frac{\mathrm{2}}{\mathrm{3}}} \:=\:\int\left(\mathrm{x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{x}−\mathrm{1}\right)\mathrm{dx} \\ $$$$\mathrm{I}\:\overset{\mathrm{u}=\mathrm{x}+\mathrm{1}} {=}\int\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{u}−\mathrm{2}\right)\mathrm{du}\:=\:\int\mathrm{u}^{\frac{\mathrm{4}}{\mathrm{3}}} \mathrm{du}\:−\:\mathrm{2}\int\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{3}}} \mathrm{du}\: \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{3u}^{\frac{\mathrm{7}}{\mathrm{3}}} }{\mathrm{7}}\:−\:\frac{\mathrm{3u}^{\frac{\mathrm{4}}{\mathrm{3}}} }{\mathrm{2}}\:+\:\mathrm{C} \\ $$
Commented by MJS_new last updated on 19/Jan/21
...=(3/(14))u^(4/3) (2u+7)=(3/(14))(x+1)^(4/3) (2x+5)+C
$$…=\frac{\mathrm{3}}{\mathrm{14}}{u}^{\mathrm{4}/\mathrm{3}} \left(\mathrm{2}{u}+\mathrm{7}\right)=\frac{\mathrm{3}}{\mathrm{14}}\left({x}+\mathrm{1}\right)^{\mathrm{4}/\mathrm{3}} \left(\mathrm{2}{x}+\mathrm{5}\right)+{C} \\ $$
Answered by Dwaipayan Shikari last updated on 19/Jan/21
∫(x−1)(x+1)^(1/3) dx  =∫(t−2)t^(1/3) dt  =(3/7)t^(7/3) −(3/2)t^(4/3) dt=(3/7)(x+1)^(7/3) −(3/2)(x+1)^(4/3) +C
$$\int\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} {dx} \\ $$$$=\int\left({t}−\mathrm{2}\right){t}^{\frac{\mathrm{1}}{\mathrm{3}}} {dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{7}}{t}^{\frac{\mathrm{7}}{\mathrm{3}}} −\frac{\mathrm{3}}{\mathrm{2}}{t}^{\frac{\mathrm{4}}{\mathrm{3}}} {dt}=\frac{\mathrm{3}}{\mathrm{7}}\left({x}+\mathrm{1}\right)^{\frac{\mathrm{7}}{\mathrm{3}}} −\frac{\mathrm{3}}{\mathrm{2}}\left({x}+\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} +{C}\: \\ $$

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