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Question Number 96746 by MJS last updated on 04/Jun/20

∫((√x)/((1+x^3 )(√(1−x^3 ))))dx=? ∫((√x)/((1−x^3 )(√(1+x^3 ))))dx=?

$$\int\frac{\sqrt{{x}}}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{3}} }}{dx}=? \\ $$$$\int\frac{\sqrt{{x}}}{\left(\mathrm{1}−{x}^{\mathrm{3}} \right)\sqrt{\mathrm{1}+{x}^{\mathrm{3}} }}{dx}=? \\ $$

Commented by bemath last updated on 04/Jun/20

I have no idea to solve this integral

Commented by MJS last updated on 05/Jun/20

thank you everybody! how to reach ∫((√x)/((1+x^3 )(√(1−x^3 ))))dx=−((√2)/3)arcsin (√((1−x^3 )/(1+x^3 ))) ?

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{everybody}! \\ $$$$\mathrm{how}\:\mathrm{to}\:\mathrm{reach} \\ $$$$\int\frac{\sqrt{{x}}}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{3}} }}{dx}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\mathrm{arcsin}\:\sqrt{\frac{\mathrm{1}−{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{3}} }}\:? \\ $$

Answered by  M±th+et+s last updated on 04/Jun/20

∫((√x)/((1+x^3 )(√(1−x^3 ))))dx let x=sin^(2/3) y dx=((2cos(y))/(3((siny))^(1/3) ))dy =∫(((sin(y)))^(1/3) /((1+sin^2 (y))cosy)).((cos(y))/((sin(y)))^(1/3) )dy =(2/3)∫(1/(1+sin^2 (y)))dy=(2/3)∫((csc^2 (y))/(csc^2 (y)+1))dy =(2/3)∫((csc^2 (y))/(cot^2 (y)+2))dy=((√2)/3)tan^(−1) ((√2)tan(y))+c =((√2)/3)tan^(−1) ((((√2)x^(3/2) )/(√(1−x^3 ))))+c

$$\int\frac{\sqrt{{x}}}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{3}} }}{dx} \\ $$$${let}\:{x}={sin}^{\frac{\mathrm{2}}{\mathrm{3}}} {y}\:\:\:\:\:\:{dx}=\frac{\mathrm{2}{cos}\left({y}\right)}{\mathrm{3}\sqrt[{\mathrm{3}}]{{siny}}}{dy} \\ $$$$=\int\frac{\sqrt[{\mathrm{3}}]{{sin}\left({y}\right)}}{\left(\mathrm{1}+{sin}^{\mathrm{2}} \left({y}\right)\right){cosy}}.\frac{{cos}\left({y}\right)}{\sqrt[{\mathrm{3}}]{{sin}\left({y}\right)}}{dy} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\int\frac{\mathrm{1}}{\mathrm{1}+{sin}^{\mathrm{2}} \left({y}\right)}{dy}=\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{csc}^{\mathrm{2}} \left({y}\right)}{{csc}^{\mathrm{2}} \left({y}\right)+\mathrm{1}}{dy} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{csc}^{\mathrm{2}} \left({y}\right)}{{cot}^{\mathrm{2}} \left({y}\right)+\mathrm{2}}{dy}=\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}{tan}\left({y}\right)\right)+{c} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}{x}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{3}} }}\right)+{c} \\ $$

Answered by Sourav mridha last updated on 04/Jun/20

(2)..let,x=tan^(2/3) 𝛂 so we get.. =(1/6)∫sec(2𝛂)d(2𝛂)+(𝛂/3)+c =(1/6)ln[tan((𝛑/4)+tan^(−1) (x^(3/2) )] +(1/3)tan^(−1) (x^(3/2) )+c 0r Ans: =ln[((1+x^(3/2) )/(1−x^(3/2) ))]^(1/6) +(1/3)tan^(−1) (x^(3/2) )+c

$$\left(\mathrm{2}\right)..\boldsymbol{{let}},\boldsymbol{{x}}=\boldsymbol{{tan}}^{\frac{\mathrm{2}}{\mathrm{3}}} \boldsymbol{\alpha}\:\boldsymbol{{so}}\:\boldsymbol{{we}}\:\boldsymbol{{get}}.. \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{6}}\int\boldsymbol{{sec}}\left(\mathrm{2}\boldsymbol{\alpha}\right)\boldsymbol{{d}}\left(\mathrm{2}\boldsymbol{\alpha}\right)+\frac{\boldsymbol{\alpha}}{\mathrm{3}}+\boldsymbol{{c}} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{6}}\boldsymbol{{ln}}\left[\mathrm{t}\boldsymbol{{an}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}+\mathrm{tan}^{−\mathrm{1}} \left(\boldsymbol{{x}}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)\right]\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \left(\boldsymbol{{x}}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)+\boldsymbol{{c}} \\ $$$$\mathrm{0}\boldsymbol{{r}}\:\boldsymbol{{Ans}}: \\ $$$$\:\:\:\:\:=\boldsymbol{{ln}}\left[\frac{\mathrm{1}+\boldsymbol{{x}}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{1}−\boldsymbol{{x}}^{\frac{\mathrm{3}}{\mathrm{2}}} }\right]^{\frac{\mathrm{1}}{\mathrm{6}}} +\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \left(\boldsymbol{{x}}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)+\boldsymbol{{c}} \\ $$

Answered by  M±th+et+s last updated on 04/Jun/20

=∫((√x)/(x^3 (x^(−3) −1)(√x^3 )(√(x^(−3) +1))))dx =(1/2)∫(x^(−4) /((x^(−3) +1)(√(x^(−3) +1))))dx =∫(1/(((x^(−3) +1)/2)−1)).(x^(−4) /(2(√(x^(−3) +1))))dx =∫(1/(1−((√((x^(−3) +1)/2)))^2 )).((−x^(−4) )/(2(√(x^(−3) +1))))dx =((√2)/3)tanh^(−1) ((√((x^(−3) +1)/2)))+c

$$=\int\frac{\sqrt{{x}}}{{x}^{\mathrm{3}} \left({x}^{−\mathrm{3}} −\mathrm{1}\right)\sqrt{{x}^{\mathrm{3}} }\sqrt{{x}^{−\mathrm{3}} +\mathrm{1}}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{−\mathrm{4}} }{\left({x}^{−\mathrm{3}} +\mathrm{1}\right)\sqrt{{x}^{−\mathrm{3}} +\mathrm{1}}}{dx} \\ $$$$=\int\frac{\mathrm{1}}{\frac{{x}^{−\mathrm{3}} +\mathrm{1}}{\mathrm{2}}−\mathrm{1}}.\frac{{x}^{−\mathrm{4}} }{\mathrm{2}\sqrt{{x}^{−\mathrm{3}} +\mathrm{1}}}{dx} \\ $$$$=\int\frac{\mathrm{1}}{\mathrm{1}−\left(\sqrt{\frac{{x}^{−\mathrm{3}} +\mathrm{1}}{\mathrm{2}}}\right)^{\mathrm{2}} }.\frac{−{x}^{−\mathrm{4}} }{\mathrm{2}\sqrt{{x}^{−\mathrm{3}} +\mathrm{1}}}{dx} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}{tanh}^{−\mathrm{1}} \left(\sqrt{\frac{{x}^{−\mathrm{3}} +\mathrm{1}}{\mathrm{2}}}\right)+{c} \\ $$

Commented by MJS last updated on 05/Jun/20

great!

$$\mathrm{great}! \\ $$

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