1-x-1-x-dx-

Question Number 168755 by MikeH last updated on 17/Apr/22

$$\int\frac{\mathrm{1}}{{x}+\sqrt{\mathrm{1}−{x}}}\:{dx} \\ $$

Commented by safojontoshtemirov last updated on 17/Apr/22

∫(1/(x+(√(1−x))))dx= (√(1−x))=t 1−x=t^2 x=1−t^2 dx=−2tdt −∫((2t)/(1−t^2 +t))dt=∫((2t)/(t^2 −t−1))dt=∫((2t−1)/(t^2 −t−1))dt+∫(1/(t^2 −t−1))dt=I_1 +I_2 I_1 =∫((d(t^2 −t−1))/(t^2 −t−1))=ln(t^2 −t−1)+C_1 =ln(−x−(√(1−x)))+C_1 x≤0 I_2 =∫(1/(t^2 −t−1))dt=∫(1/((t−(1/2))^2 −(((√5)/2))^2 ))dt ∫(1/((t−(1/2)−((√5)/2))(t−(1/2)+((√5)/2))))dt=(1/( (√5)))∫((1/(t−(1/2)−((√5)/2)))−(1/(t−(1/2)+((√5)/2))))dt I_2 =(1/( (√5)))ln(t−(1/2)−((√5)/2))−(1/( (√5)))ln(t−(1/2)+((√5)/2))+C_2 ln(−x−(√(1−x)))+(1/( (√5)))ln(((2(√(1−x))−1−(√5))/(2(√(1−x))−1+(√5))))+C

$$\int\frac{\mathrm{1}}{{x}+\sqrt{\mathrm{1}−{x}}}{dx}=\:\:\:\sqrt{\mathrm{1}−{x}}={t}\:\:\:\mathrm{1}−{x}={t}^{\mathrm{2}} \:\:{x}=\mathrm{1}−{t}^{\mathrm{2}} \:{dx}=−\mathrm{2}{tdt} \\ $$$$−\int\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} +{t}}{dt}=\int\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} −{t}−\mathrm{1}}{dt}=\int\frac{\mathrm{2}{t}−\mathrm{1}}{{t}^{\mathrm{2}} −{t}−\mathrm{1}}{dt}+\int\frac{\mathrm{1}}{{t}^{\mathrm{2}} −{t}−\mathrm{1}}{dt}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =\int\frac{{d}\left({t}^{\mathrm{2}} −{t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} −{t}−\mathrm{1}}={ln}\left({t}^{\mathrm{2}} −{t}−\mathrm{1}\right)+{C}_{\mathrm{1}} ={ln}\left(−{x}−\sqrt{\mathrm{1}−{x}}\right)+{C}_{\mathrm{1}} \:\:{x}\leqslant\mathrm{0} \\ $$$${I}_{\mathrm{2}} =\int\frac{\mathrm{1}}{{t}^{\mathrm{2}} −{t}−\mathrm{1}}{dt}=\int\frac{\mathrm{1}}{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} }{dt} \\ $$$$\int\frac{\mathrm{1}}{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left({t}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)}{dt}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\int\left(\frac{\mathrm{1}}{{t}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}}−\frac{\mathrm{1}}{{t}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}}\right){dt} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}{ln}\left({t}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}{ln}\left({t}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)+{C}_{\mathrm{2}} \\ $$$${ln}\left(−{x}−\sqrt{\mathrm{1}−{x}}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}{ln}\left(\frac{\mathrm{2}\sqrt{\mathrm{1}−{x}}−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{1}−{x}}−\mathrm{1}+\sqrt{\mathrm{5}}}\right)+{C} \\ $$$$ \\ $$

Commented by Dildora last updated on 17/Apr/22