Question-226609

Question Number 226609 by Spillover last updated on 07/Dec/25

Answered by Ghisom_ last updated on 08/Dec/25

∫_(−2) ^(−1) ((√(2+x))/(x(√(2−x))))dx= [t=((√(2+x))/( (√(2−x)))) → dx=((√((2−x)^3 (2+x)))/2)dt] =4∫_0 ^(1/(√3)) (t^2 /((t−1)(t+1)(t^2 +1)))dt= =∫_0 ^(1/(√3)) (dt/(t−1))−∫_0 ^(1/(√3)) (dt/(t+1))+2∫_0 ^(1/(√3)) (dt/(t^2 +1))= =[ln ∣t−1∣ −ln ∣t+1∣ +2arctan t]_0 ^(1/(√3)) = =(π/3)+ln (2−(√3))

$$\underset{−\mathrm{2}} {\overset{−\mathrm{1}} {\int}}\frac{\sqrt{\mathrm{2}+{x}}}{{x}\sqrt{\mathrm{2}−{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt{\mathrm{2}+{x}}}{\:\sqrt{\mathrm{2}−{x}}}\:\rightarrow\:{dx}=\frac{\sqrt{\left(\mathrm{2}−{x}\right)^{\mathrm{3}} \left(\mathrm{2}+{x}\right)}}{\mathrm{2}}{dt}\right] \\ $$$$=\mathrm{4}\underset{\mathrm{0}} {\overset{\mathrm{1}/\sqrt{\mathrm{3}}} {\int}}\frac{{t}^{\mathrm{2}} }{\left({t}−\mathrm{1}\right)\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}= \\ $$$$=\underset{\mathrm{0}} {\overset{\mathrm{1}/\sqrt{\mathrm{3}}} {\int}}\frac{{dt}}{{t}−\mathrm{1}}−\underset{\mathrm{0}} {\overset{\mathrm{1}/\sqrt{\mathrm{3}}} {\int}}\frac{{dt}}{{t}+\mathrm{1}}+\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}/\sqrt{\mathrm{3}}} {\int}}\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\left[\mathrm{ln}\:\mid{t}−\mathrm{1}\mid\:−\mathrm{ln}\:\mid{t}+\mathrm{1}\mid\:+\mathrm{2arctan}\:{t}\right]_{\mathrm{0}} ^{\mathrm{1}/\sqrt{\mathrm{3}}} = \\ $$$$=\frac{\pi}{\mathrm{3}}+\mathrm{ln}\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\right) \\ $$

Commented by Spillover last updated on 09/Dec/25