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Question-198420




Question Number 198420 by Mingma last updated on 19/Oct/23
Answered by Frix last updated on 19/Oct/23
∫(x^2 /((3+4x−4x^2 )^(3/2) ))dx =_(dx=−(((2x−1)(√(3+4x−4x^2 )))/(4t))) ^(t=((2+(√(3+4x−4x^2 )))/(2x−1)))    =−(1/(16))∫(((t^2 +4t+1)^2 )/((t−1)^2 (t+1)^2 (t^2 +1)))dt=  =∫(−(9/(32(t−1)^2 ))−(1/(32(t+1)^2 ))+(1/(4(t^2 +1))))dt=  =(9/(32(t−1)))+(1/(32(t+1)))+((tan^(−1)  t)/4)=  =((5t+4)/(16(t^2 −1)))+((tan^(−1)  t)/4)=  =((10x+3)/(32(√(3+4x−4x^2 ))))+((tan^(−1)  ((2+(√(3+4x−4x^2 )))/(2x−1)))/4)+C
$$\int\frac{{x}^{\mathrm{2}} }{\left(\mathrm{3}+\mathrm{4}{x}−\mathrm{4}{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx}\:\underset{{dx}=−\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{\mathrm{3}+\mathrm{4}{x}−\mathrm{4}{x}^{\mathrm{2}} }}{\mathrm{4}{t}}} {\overset{{t}=\frac{\mathrm{2}+\sqrt{\mathrm{3}+\mathrm{4}{x}−\mathrm{4}{x}^{\mathrm{2}} }}{\mathrm{2}{x}−\mathrm{1}}} {=}}\: \\ $$$$=−\frac{\mathrm{1}}{\mathrm{16}}\int\frac{\left({t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{1}\right)^{\mathrm{2}} }{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \left({t}+\mathrm{1}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}= \\ $$$$=\int\left(−\frac{\mathrm{9}}{\mathrm{32}\left({t}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{32}\left({t}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}\right){dt}= \\ $$$$=\frac{\mathrm{9}}{\mathrm{32}\left({t}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{32}\left({t}+\mathrm{1}\right)}+\frac{\mathrm{tan}^{−\mathrm{1}} \:{t}}{\mathrm{4}}= \\ $$$$=\frac{\mathrm{5}{t}+\mathrm{4}}{\mathrm{16}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\mathrm{tan}^{−\mathrm{1}} \:{t}}{\mathrm{4}}= \\ $$$$=\frac{\mathrm{10}{x}+\mathrm{3}}{\mathrm{32}\sqrt{\mathrm{3}+\mathrm{4}{x}−\mathrm{4}{x}^{\mathrm{2}} }}+\frac{\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{2}+\sqrt{\mathrm{3}+\mathrm{4}{x}−\mathrm{4}{x}^{\mathrm{2}} }}{\mathrm{2}{x}−\mathrm{1}}}{\mathrm{4}}+{C} \\ $$

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