Question Number 187317 by ajfour last updated on 15/Feb/23
$$\int\frac{{dx}}{{x}\sqrt{\mathrm{1}−\mathrm{2}{x}}} \\ $$
Answered by MJS_new last updated on 15/Feb/23
![∫(dx/(x(√(1−2x))))= [t=(√(1−2x)) → dx=−(√(1−2x))dt] =2∫(dt/(t^2 −1))=ln ((t−1)/(t+1)) = =ln ∣((1−(√(1−2x)))/(1+(√(1−2x))))∣ +C](https://www.tinkutara.com/question/Q187319.png)
$$\int\frac{{dx}}{{x}\sqrt{\mathrm{1}−\mathrm{2}{x}}}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{1}−\mathrm{2}{x}}\:\rightarrow\:{dx}=−\sqrt{\mathrm{1}−\mathrm{2}{x}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{1}}=\mathrm{ln}\:\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\:= \\ $$$$=\mathrm{ln}\:\mid\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{2}{x}}}{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{2}{x}}}\mid\:+{C} \\ $$
Commented by ajfour last updated on 16/Feb/23
$${thank}\:{you},\:{i}\:{shall}\:{make}\:{good}\:{use} \\ $$$${of}\:{it}! \\ $$
Answered by Humble last updated on 16/Feb/23
![let u = 1−2x ∫(1/( (√u)(u−1)))du let s=(√u) ==> u=s^2 ds =(1/(2(√u)))du ==> du=2(√(u ))ds ∫((2(√u))/( (√u)(s^2 −1)))ds ∫(2/((s^2 −1)))ds 2∫(1/((s^2 −1)))ds ==> 2[−(((ln∣s+1∣)/2)−((ln∣s−1∣)/2))] s=(√u)==> u=(√(1−2x)) −ln∣(√(1−2x))+1∣ +ln∣(√(1−2x))−1∣ ln∣(√(1−2x))−1∣ − ln∣(√(1−2x))+1∣+C](https://www.tinkutara.com/question/Q187334.png)
$${let}\:{u}\:=\:\mathrm{1}−\mathrm{2}{x} \\ $$$$\int\frac{\mathrm{1}}{\:\sqrt{{u}}\left({u}−\mathrm{1}\right)}{du} \\ $$$${let}\:{s}=\sqrt{{u}}\:==>\:{u}={s}^{\mathrm{2}} \\ $$$${ds}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{u}}}{du}\:==>\:{du}=\mathrm{2}\sqrt{{u}\:}{ds} \\ $$$$\int\frac{\mathrm{2}\sqrt{{u}}}{\:\sqrt{{u}}\left({s}^{\mathrm{2}} −\mathrm{1}\right)}{ds} \\ $$$$\int\frac{\mathrm{2}}{\left({s}^{\mathrm{2}} −\mathrm{1}\right)}{ds} \\ $$$$\mathrm{2}\int\frac{\mathrm{1}}{\left({s}^{\mathrm{2}} −\mathrm{1}\right)}{ds}\:\:==>\:\mathrm{2}\left[−\left(\frac{{ln}\mid{s}+\mathrm{1}\mid}{\mathrm{2}}−\frac{{ln}\mid{s}−\mathrm{1}\mid}{\mathrm{2}}\right)\right] \\ $$$${s}=\sqrt{{u}}==>\:{u}=\sqrt{\mathrm{1}−\mathrm{2}{x}} \\ $$$$−{ln}\mid\sqrt{\mathrm{1}−\mathrm{2}{x}}+\mathrm{1}\mid\:+{ln}\mid\sqrt{\mathrm{1}−\mathrm{2}{x}}−\mathrm{1}\mid \\ $$$${ln}\mid\sqrt{\mathrm{1}−\mathrm{2}{x}}−\mathrm{1}\mid\:−\:{ln}\mid\sqrt{\mathrm{1}−\mathrm{2}{x}}+\mathrm{1}\mid+{C} \\ $$$$ \\ $$