Question Number 73275 by solihin last updated on 09/Nov/19
$$ \\ $$$$ \\ $$$$\int\frac{\mathrm{4}}{{x}^{\mathrm{2}} \sqrt{\mathrm{4}−{x}\delta\varkappa}}\:\:\:\:? \\ $$$$ \\ $$
Commented by MJS last updated on 09/Nov/19
$$\mathrm{the}\:\mathrm{syntax}\:\mathrm{is}\:\mathrm{not}\:\mathrm{correct} \\ $$
Commented by MJS last updated on 09/Nov/19
![∫(4/(x^2 (√(4−x))))dx= [t=(√(4−x)) → dx=−2(√(4−x))dt] =−8∫(dt/((t^2 −4)^2 ))= =−(1/2)∫(dt/((t−2)^2 ))+(1/4)∫(dt/(t−2))−(1/2)∫(dt/((t+2)^2 ))−(1/4)∫(dt/(t+2))= =(1/(2(t−2)))+(1/4)ln (t−2) +(1/(2(t+2)))−(1/4)ln (t+2) = =(t/(t^2 −4))+(1/4)ln ((t−2)/(t+2)) = =−((√(4−x))/x)+(1/4)ln ∣((x−8+4(√(4−x)))/x)∣ +C](Q73286.png)
$$\int\frac{\mathrm{4}}{{x}^{\mathrm{2}} \sqrt{\mathrm{4}−{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{4}−{x}}\:\rightarrow\:{dx}=−\mathrm{2}\sqrt{\mathrm{4}−{x}}{dt}\right] \\ $$$$=−\mathrm{8}\int\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{2}} }= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\left({t}−\mathrm{2}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dt}}{{t}−\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\left({t}+\mathrm{2}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dt}}{{t}+\mathrm{2}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({t}−\mathrm{2}\right)}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\left({t}−\mathrm{2}\right)\:+\frac{\mathrm{1}}{\mathrm{2}\left({t}+\mathrm{2}\right)}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\left({t}+\mathrm{2}\right)\:= \\ $$$$=\frac{{t}}{{t}^{\mathrm{2}} −\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\frac{{t}−\mathrm{2}}{{t}+\mathrm{2}}\:= \\ $$$$=−\frac{\sqrt{\mathrm{4}−{x}}}{{x}}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid\frac{{x}−\mathrm{8}+\mathrm{4}\sqrt{\mathrm{4}−{x}}}{{x}}\mid\:+{C} \\ $$