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Question Number 7506 by gourav~ last updated on 01/Sep/16
∫{((1−(√x))/(1+(√x)))}^(1/2) (dx/x)=?
$$\int\left\{\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}+\sqrt{{x}}}\right\}^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{dx}}{{x}}=? \\ $$$$ \\ $$$$ \\ $$
Answered by Yozzia last updated on 01/Sep/16
Let I=∫(1/x)(√((1−(√x))/(1+(√x))))dx. Let u=(((1−(√x))/(1+(√x))))^(1/2) ⇒u^2 +u^2 (√x)=1−(√x) (√x)(u^2 +1)=1−u^2 (√x)=((1−u^2 )/(1+u^2 ))⇒x=(((1−u^2 )/(1+u^2 )))^2 dx=2×(((1+u^2 )(−2u)−(1−u^2 )(2u))/((1+u^2 )^2 ))×((1−u^2 )/((1+u^2 )))du dx=((−8u(1−u^2 ))/((1+u^2 )^3 ))du ∴ I=∫(((1+u^2 )^2 )/((1−u^2 )^2 ))×u×((−8u(1−u^2 ))/((1+u^2 )^3 ))du I=∫((−8u^2 )/((1−u^2 )(1+u^2 )))du I=∫((−8u^2 )/((1+u)(1−u)(1+u^2 )))du. I=−8∫(u^2 /((1−u^2 )(1+u^2 )))du I=−8∫(1/(1−u^2 ))(1−(1/(1+u^2 )))du I=−8∫{(1/(1−u^2 ))+(1/((u^2 −1)(u^2 +1)))}du −−−−−−−−−−−−−−−−−−−−−−−−−−−− (1/(u^2 −1))−(1/(u^2 +1))=(2/((u^2 −1)(u^2 +1))) ⇒(1/((u^2 −1)(u^2 +1)))=(1/2)((1/((u−1)(u+1)))−(1/(u^2 +1))) ⇒(1/((u^2 −1)(u^2 +1)))=(1/4)((1/(u−1))−(1/(u+1)))−(1/(2(u^2 +1))) −−−−−−−−−−−−−−−−−−−−−−−−−−−− I=−8∫{(1/2)((1/(1+u))+(1/(1−u)))+(1/4)(−(1/(1−u))−(1/(1+u)))−(1/(2(u^2 +1)))}du I=−8∫{(1/4)((1/(1+u))+(1/(1−u)))−(1/(2(u^2 +1)))}du I=−8{(1/4)ln∣((1+u)/(1−u))∣−(1/2)tan^(−1) u}+C I=−2ln∣((1+u)/(1−u))∣+4tan^(−1) u+C ∫(1/x)(√((1−(√x))/(1+(√x))))dx=4tan^(−1) (√((1−(√x))/(1+(√x))))−2ln∣((1+(√((1−(√x))/(1+(√x)))))/(1−(√((1−(√x))/(1+(√x))))))∣+C ∫(1/x)(√((1−(√x))/(1+(√x))))dx=4tan^(−1) (√((1−(√x))/(1+(√x))))−2ln∣(((√(1+(√x)))+(√(1−(√x))))/( (√(1+(√x)))−(√(1−(√x)))))∣+C ∫(1/x)(√((1−(√x))/(1+(√x))))dx=4tan^(−1) (√((1−(√x))/(1+(√x))))−2ln∣((1+(√(1−x)))/( (√x)))∣+C
$${Let}\:{I}=\int\frac{\mathrm{1}}{{x}}\sqrt{\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}+\sqrt{{x}}}}{dx}. \\ $$$${Let}\:{u}=\left(\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}+\sqrt{{x}}}\right)^{\mathrm{1}/\mathrm{2}} \Rightarrow{u}^{\mathrm{2}} +{u}^{\mathrm{2}} \sqrt{{x}}=\mathrm{1}−\sqrt{{x}} \\ $$$$\sqrt{{x}}\left({u}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{1}−{u}^{\mathrm{2}} \\ $$$$\sqrt{{x}}=\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\Rightarrow{x}=\left(\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$${dx}=\mathrm{2}×\frac{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(−\mathrm{2}{u}\right)−\left(\mathrm{1}−{u}^{\mathrm{2}} \right)\left(\mathrm{2}{u}\right)}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }×\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{du} \\ $$$${dx}=\frac{−\mathrm{8}{u}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{3}} }{du} \\ $$$$\therefore\:{I}=\int\frac{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\mathrm{2}} }×{u}×\frac{−\mathrm{8}{u}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{3}} }{du} \\ $$$${I}=\int\frac{−\mathrm{8}{u}^{\mathrm{2}} }{\left(\mathrm{1}−{u}^{\mathrm{2}} \right)\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{du} \\ $$$${I}=\int\frac{−\mathrm{8}{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}\right)\left(\mathrm{1}−{u}\right)\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{du}. \\ $$$${I}=−\mathrm{8}\int\frac{{u}^{\mathrm{2}} }{\left(\mathrm{1}−{u}^{\mathrm{2}} \right)\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{du} \\ $$$${I}=−\mathrm{8}\int\frac{\mathrm{1}}{\mathrm{1}−{u}^{\mathrm{2}} }\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\right){du} \\ $$$${I}=−\mathrm{8}\int\left\{\frac{\mathrm{1}}{\mathrm{1}−{u}^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({u}^{\mathrm{2}} −\mathrm{1}\right)\left({u}^{\mathrm{2}} +\mathrm{1}\right)}\right\}{du} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$$\frac{\mathrm{1}}{{u}^{\mathrm{2}} −\mathrm{1}}−\frac{\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{2}}{\left({u}^{\mathrm{2}} −\mathrm{1}\right)\left({u}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\left({u}^{\mathrm{2}} −\mathrm{1}\right)\left({u}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\left({u}−\mathrm{1}\right)\left({u}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{\left({u}^{\mathrm{2}} −\mathrm{1}\right)\left({u}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{{u}−\mathrm{1}}−\frac{\mathrm{1}}{{u}+\mathrm{1}}\right)−\frac{\mathrm{1}}{\mathrm{2}\left({u}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${I}=−\mathrm{8}\int\left\{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}+{u}}+\frac{\mathrm{1}}{\mathrm{1}−{u}}\right)+\frac{\mathrm{1}}{\mathrm{4}}\left(−\frac{\mathrm{1}}{\mathrm{1}−{u}}−\frac{\mathrm{1}}{\mathrm{1}+{u}}\right)−\frac{\mathrm{1}}{\mathrm{2}\left({u}^{\mathrm{2}} +\mathrm{1}\right)}\right\}{du} \\ $$$${I}=−\mathrm{8}\int\left\{\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{1}+{u}}+\frac{\mathrm{1}}{\mathrm{1}−{u}}\right)−\frac{\mathrm{1}}{\mathrm{2}\left({u}^{\mathrm{2}} +\mathrm{1}\right)}\right\}{du} \\ $$$${I}=−\mathrm{8}\left\{\frac{\mathrm{1}}{\mathrm{4}}{ln}\mid\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\mid−\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} {u}\right\}+{C} \\ $$$${I}=−\mathrm{2}{ln}\mid\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\mid+\mathrm{4}{tan}^{−\mathrm{1}} {u}+{C} \\ $$$$\int\frac{\mathrm{1}}{{x}}\sqrt{\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}+\sqrt{{x}}}}{dx}=\mathrm{4}{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}+\sqrt{{x}}}}−\mathrm{2}{ln}\mid\frac{\mathrm{1}+\sqrt{\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}+\sqrt{{x}}}}}{\mathrm{1}−\sqrt{\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}+\sqrt{{x}}}}}\mid+{C} \\ $$$$\int\frac{\mathrm{1}}{{x}}\sqrt{\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}+\sqrt{{x}}}}{dx}=\mathrm{4}{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}+\sqrt{{x}}}}−\mathrm{2}{ln}\mid\frac{\sqrt{\mathrm{1}+\sqrt{{x}}}+\sqrt{\mathrm{1}−\sqrt{{x}}}}{\:\sqrt{\mathrm{1}+\sqrt{{x}}}−\sqrt{\mathrm{1}−\sqrt{{x}}}}\mid+{C} \\ $$$$\int\frac{\mathrm{1}}{{x}}\sqrt{\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}+\sqrt{{x}}}}{dx}=\mathrm{4}{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}+\sqrt{{x}}}}−\mathrm{2}{ln}\mid\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}}}{\:\sqrt{{x}}}\mid+{C} \\ $$$$ \\ $$$$ \\ $$

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