Question Number 219586 by Hery03 last updated on 29/Apr/25
$${Integrate}\:: \\ $$$$\int\frac{{x}^{\mathrm{2}} }{\:\sqrt{{x}^{\mathrm{2}} \:−\:{x}}\:+\:\mathrm{1}}{dx}. \\ $$
Answered by Ghisom last updated on 29/Apr/25
![∫(x^2 /( 1+(√x)(√(x−1))))dx= [t=((√(x−1))/( (√x))) → dx=((2t)/((1−t^2 )^2 ))dt] [x=(1/(1−t^2 )) ⇒ (√x)(√(x−1))=(t/(1−t^2 ))] =2∫(t/((t^2 −1)^3 (t^2 −t−1)))dt= =Σ_(k=1) ^8 I_k I_1 =((3/( (√5)))−1)∫(dt/(t−((1+(√5))/2))) I_2 =−((3/( (√5)))+1)∫(dt/(t−((1−(√5))/2))) I_3 =−(3/8)∫(dt/(t−1)) I_4 =((19)/8)∫(dt/(t+1)) I_5 =−(1/8)∫(dt/((t−1)^2 )) I_6 =(7/8)∫(dt/((t+1)^2 )) I_7 =−(1/4)∫(dt/((t−1)^3 )) I_8 =(1/4)∫(dt/((t+1)^3 )) these are easy to solve](https://www.tinkutara.com/question/Q219592.png)
$$\int\frac{{x}^{\mathrm{2}} }{\:\mathrm{1}+\sqrt{{x}}\sqrt{{x}−\mathrm{1}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}}}\:\rightarrow\:{dx}=\frac{\mathrm{2}{t}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\right] \\ $$$$\:\:\:\:\:\left[{x}=\frac{\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{2}} }\:\Rightarrow\:\sqrt{{x}}\sqrt{{x}−\mathrm{1}}=\frac{{t}}{\mathrm{1}−{t}^{\mathrm{2}} }\right] \\ $$$$=\mathrm{2}\int\frac{{t}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} \left({t}^{\mathrm{2}} −{t}−\mathrm{1}\right)}{dt}= \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{\mathrm{8}} {\sum}}{I}_{{k}} \\ $$$${I}_{\mathrm{1}} =\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{5}}}−\mathrm{1}\right)\int\frac{{dt}}{{t}−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}} \\ $$$${I}_{\mathrm{2}} =−\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{5}}}+\mathrm{1}\right)\int\frac{{dt}}{{t}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}} \\ $$$${I}_{\mathrm{3}} =−\frac{\mathrm{3}}{\mathrm{8}}\int\frac{{dt}}{{t}−\mathrm{1}} \\ $$$${I}_{\mathrm{4}} =\frac{\mathrm{19}}{\mathrm{8}}\int\frac{{dt}}{{t}+\mathrm{1}} \\ $$$${I}_{\mathrm{5}} =−\frac{\mathrm{1}}{\mathrm{8}}\int\frac{{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${I}_{\mathrm{6}} =\frac{\mathrm{7}}{\mathrm{8}}\int\frac{{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${I}_{\mathrm{7}} =−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${I}_{\mathrm{8}} =\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\mathrm{these}\:\mathrm{are}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$