Question Number 220644 by Nicholas666 last updated on 17/May/25
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{1}} ^{\:\mathrm{2}} \:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\:\sqrt{\left(\mathrm{2}{x}\:−\:\mathrm{1}\right)\centerdot\left(\mathrm{2}{x}\:+\:\mathrm{2}\right)}}\:{dx} \\ $$$$ \\ $$
Answered by Ghisom last updated on 17/May/25
![2∫_1 ^2 (x^2 /( (√(2x−1))(√(2x+2))))dx= [t=2x] =(1/4)∫_2 ^4 (t^2 /( (√(t−1))(√(t+2))))dt= [u=(√(t−1))+(√(t+2)) → dt=((2(√(t−1))(√(t+2)))/u)du] =(1/(32))∫_3 ^((√6)+(√3)) (((u^4 −2u^2 +9)^2 )/u^5 )du= =∫_3 ^((√6)+(√3)) ((u^3 /(32))−(u/8)+((11)/(16u))−(9/(8u^3 ))+((81)/(32u^5 )))du= ... =[((11)/(16))ln u +(((u^2 −3)(u^2 +3)(u^4 −8u^2 +9))/(128u^4 ))]_3 ^((√6)+(√3)) = =((11)/(16))ln ((1+(√2))/( (√3))) +((15(√2))/(16))−(1/8) 2∫(x^2 /( (√(2x−1))(√(2x+2))))dx= =((11)/(16))ln ((√(2x−1))+(√(2x+2)))+(((4x−3)(√(2x−1))(√(2x+2)))/(16))+C](https://www.tinkutara.com/question/Q220654.png)
$$\mathrm{2}\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{2}{x}−\mathrm{1}}\sqrt{\mathrm{2}{x}+\mathrm{2}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{2}{x}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\frac{{t}^{\mathrm{2}} }{\:\sqrt{{t}−\mathrm{1}}\sqrt{{t}+\mathrm{2}}}{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\sqrt{{t}−\mathrm{1}}+\sqrt{{t}+\mathrm{2}}\:\rightarrow\:{dt}=\frac{\mathrm{2}\sqrt{{t}−\mathrm{1}}\sqrt{{t}+\mathrm{2}}}{{u}}{du}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{32}}\underset{\mathrm{3}} {\overset{\sqrt{\mathrm{6}}+\sqrt{\mathrm{3}}} {\int}}\frac{\left({u}^{\mathrm{4}} −\mathrm{2}{u}^{\mathrm{2}} +\mathrm{9}\right)^{\mathrm{2}} }{{u}^{\mathrm{5}} }{du}= \\ $$$$=\underset{\mathrm{3}} {\overset{\sqrt{\mathrm{6}}+\sqrt{\mathrm{3}}} {\int}}\left(\frac{{u}^{\mathrm{3}} }{\mathrm{32}}−\frac{{u}}{\mathrm{8}}+\frac{\mathrm{11}}{\mathrm{16}{u}}−\frac{\mathrm{9}}{\mathrm{8}{u}^{\mathrm{3}} }+\frac{\mathrm{81}}{\mathrm{32}{u}^{\mathrm{5}} }\right){du}= \\ $$$$… \\ $$$$=\left[\frac{\mathrm{11}}{\mathrm{16}}\mathrm{ln}\:{u}\:+\frac{\left({u}^{\mathrm{2}} −\mathrm{3}\right)\left({u}^{\mathrm{2}} +\mathrm{3}\right)\left({u}^{\mathrm{4}} −\mathrm{8}{u}^{\mathrm{2}} +\mathrm{9}\right)}{\mathrm{128}{u}^{\mathrm{4}} }\right]_{\mathrm{3}} ^{\sqrt{\mathrm{6}}+\sqrt{\mathrm{3}}} = \\ $$$$=\frac{\mathrm{11}}{\mathrm{16}}\mathrm{ln}\:\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\:+\frac{\mathrm{15}\sqrt{\mathrm{2}}}{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$ \\ $$$$\mathrm{2}\int\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{2}{x}−\mathrm{1}}\sqrt{\mathrm{2}{x}+\mathrm{2}}}{dx}= \\ $$$$=\frac{\mathrm{11}}{\mathrm{16}}\mathrm{ln}\:\left(\sqrt{\mathrm{2}{x}−\mathrm{1}}+\sqrt{\mathrm{2}{x}+\mathrm{2}}\right)+\frac{\left(\mathrm{4}{x}−\mathrm{3}\right)\sqrt{\mathrm{2}{x}−\mathrm{1}}\sqrt{\mathrm{2}{x}+\mathrm{2}}}{\mathrm{16}}+{C} \\ $$
Answered by Frix last updated on 17/May/25
![2∫_1 ^2 ((2x^2 )/( (√(2x−1))(√(2x+2))))dx =^([t=((√(2x−1))/( (√(2x+2))))]) =−(1/2)∫_(1/2) ^((√2)/2) (((2t^2 +1)^2 )/((t^2 −1)^3 ))dt =_(Method]) ^([Ostrogradski′s) =[((3t(7t^2 −1))/(16(t^2 −1)^2 ))]_(1/2) ^((√2)/2) −((11)/(16))∫_(1/2) ^((√2)/2) (dt/(t^2 −1))= =((15(√2))/(16))−(1/8)+[((11)/(32))ln ∣((t+1)/(t−1))∣]_(1/2) ^((√2)/2) = =((30(√2)−4+22ln (1+(√2)) −11ln 3)/(32))](https://www.tinkutara.com/question/Q220675.png)
$$\mathrm{2}\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\frac{\mathrm{2}{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{2}{x}−\mathrm{1}}\sqrt{\mathrm{2}{x}+\mathrm{2}}}{dx}\:\overset{\left[{t}=\frac{\sqrt{\mathrm{2}{x}−\mathrm{1}}}{\:\sqrt{\mathrm{2}{x}+\mathrm{2}}}\right]} {=} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\underset{\frac{\mathrm{1}}{\mathrm{2}}} {\overset{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} {\int}}\frac{\left(\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} }{dt}\:\underset{\left.\mathrm{Method}\right]} {\overset{\left[\mathrm{Ostrogradski}'\mathrm{s}\right.} {=}} \\ $$$$=\left[\frac{\mathrm{3}{t}\left(\mathrm{7}{t}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{16}\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\right]_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} −\frac{\mathrm{11}}{\mathrm{16}}\underset{\frac{\mathrm{1}}{\mathrm{2}}} {\overset{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} {\int}}\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{1}}= \\ $$$$=\frac{\mathrm{15}\sqrt{\mathrm{2}}}{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{8}}+\left[\frac{\mathrm{11}}{\mathrm{32}}\mathrm{ln}\:\mid\frac{{t}+\mathrm{1}}{{t}−\mathrm{1}}\mid\right]_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} = \\ $$$$=\frac{\mathrm{30}\sqrt{\mathrm{2}}−\mathrm{4}+\mathrm{22ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:−\mathrm{11ln}\:\mathrm{3}}{\mathrm{32}} \\ $$