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Question Number 11148 by suci last updated on 14/Mar/17

∫((x dx)/((x+1))^(1/3) )=....???

$$\int\frac{{x}\:{dx}}{\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}}=....??? \\ $$

Answered by ajfour last updated on 14/Mar/17

= ((3(x+1)^(5/3) )/5) −((3(x+1)^(2/3) )/2) +C

$$=\:\frac{\mathrm{3}\left({x}+\mathrm{1}\right)^{\mathrm{5}/\mathrm{3}} }{\mathrm{5}}\:−\frac{\mathrm{3}\left({x}+\mathrm{1}\right)^{\mathrm{2}/\mathrm{3}} }{\mathrm{2}}\:+{C}\: \\ $$

Commented by ajfour last updated on 14/Mar/17

let x+1=t  x=t−1   ;  dx=dt  so,  ∫(x/((x+1)^(1/3) )) dx = ∫(t−1)t^(−1/3)  dt  =∫(t^(2/3)  −t^(−1/3) )dt  ((3t^(5/3) )/5) − ((3t^(2/3) )/2) +C

$${let}\:{x}+\mathrm{1}={t} \\ $$$${x}={t}−\mathrm{1}\:\:\:;\:\:{dx}={dt} \\ $$$${so},\:\:\int\frac{{x}}{\left({x}+\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} }\:{dx}\:=\:\int\left({t}−\mathrm{1}\right){t}^{−\mathrm{1}/\mathrm{3}} \:{dt} \\ $$$$=\int\left({t}^{\mathrm{2}/\mathrm{3}} \:−{t}^{−\mathrm{1}/\mathrm{3}} \right){dt} \\ $$$$\frac{\mathrm{3}{t}^{\mathrm{5}/\mathrm{3}} }{\mathrm{5}}\:−\:\frac{\mathrm{3}{t}^{\mathrm{2}/\mathrm{3}} }{\mathrm{2}}\:+{C}\:\: \\ $$

Commented by chux last updated on 17/Mar/17

x+y+z=1.....(1)  x^2 +y^2 +z^2 =35.....(2)  x^3 +y^3 +z^3 =97.....(3)        solve the equation

$$\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{1}.....\left(\mathrm{1}\right) \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\mathrm{35}.....\left(\mathrm{2}\right) \\ $$$$\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} +\mathrm{z}^{\mathrm{3}} =\mathrm{97}.....\left(\mathrm{3}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{solve}\:\mathrm{the}\:\mathrm{equation} \\ $$

Commented by ajfour last updated on 17/Mar/17

solve   x^3 −x^2 −17x−15=0  roots are −1,−3, 5  x=−1  y=−3  z=5

$${solve}\:\:\:{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{17}{x}−\mathrm{15}=\mathrm{0} \\ $$$$\mathrm{roots}\:\mathrm{are}\:−\mathrm{1},−\mathrm{3},\:\mathrm{5} \\ $$$$\mathrm{x}=−\mathrm{1} \\ $$$$\mathrm{y}=−\mathrm{3} \\ $$$$\mathrm{z}=\mathrm{5} \\ $$

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