Question Number 208733 by Frix last updated on 22/Jun/24
$$\mathrm{2}\underset{\frac{\mathrm{1}}{\mathrm{3}}} {\overset{\mathrm{1}} {\int}}\frac{{x}\sqrt{−\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{1}}}{\mathrm{7}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}}{dx}=? \\ $$$$\mathrm{Exact}\:\mathrm{solution}\:\mathrm{needed}. \\ $$
Answered by Ghisom last updated on 24/Jun/24
![2∫_(1/3) ^1 ((x(√(−3x^2 +4x−1)))/(7x^2 −4x+1))dx= [t=arcsin (3x−2) → dx=((√(−3x^2 +4x−1))/( (√3)))dt] =((2(√3))/( 3))∫_(−π/2) ^(π/2) (((2+sin t)cos^2 t)/(7sin^2 t +16sin t +13))dt= =((24(√3))/(49))∫_(−π/2) ^(π/2) ((2+3sin t)/(7sin^2 t +16sin t +13))dt+((2(√3))/(147))∫_(−π/2) ^(π/2) (2−7sin t)dt ((2(√3))/(147))∫_(−π/2) ^(π/2) (2−7sin t)dt=((2(√3))/(147))[2t+7cos t]_(−π/2) ^(π/2) = =((4(√3)π)/(147)) ((24(√3))/(49))∫_(−π/2) ^(π/2) ((2+3sin t)/(7sin^2 t +16sin t +13))dt= [u=tan (t/2) → dt=((2du)/(u^2 +1))] =((96(√3))/(49))∫_(−1) ^1 ((u^2 +3u+1)/(13u^4 +32u^3 +54u^2 +32u+13))du= =((96(√3))/(637))∫_(−1) ^1 ((u^2 +3u+1)/(Π(u^2 +((4(4±(√3)))/(13))u+((19±8(√3))/(13)))))du= =I_1 +I_2 I_1 =((16)/(49))∫_(−1) ^1 ((u+((29−4(√3))/(52)))/(u^2 +((4(4−(√3)))/(13))u+((19−8(√3))/(13))))du I_2 =−((16)/(49))∫_(−1) ^1 ((u+((29+4(√3))/(52)))/(u^2 +((4(4+(√3)))/(13))u+((19+8(√3))/(13))))du I_1 =[((4(√3))/(147))arctan (((4+(√3))u+2)/3) +(8/(49))ln (u^2 +((4(4−(√3)))/(13))u+((19−8(√3))/(13)))]_(−1) ^1 I_2 =[((4(√3))/(147))arctan (((4−(√3))u+2)/3) −(8/(49))ln (u^2 +((4(4+(√3)))/(13))u+((19+8(√3))/(13)))]_(−1) ^1 I_1 +I_2 =((4(√3))/(147))(arctan ((6+(√3))/3) +ectan ((6−(√3))/3) +arctan ((2+(√3))/3) +arctan ((2−(√3))/3))= =((4(√3)π)/(147)) ⇒ 2∫_(1/3) ^1 ((x(√(−3x^2 +4x−1)))/(7x^2 −4x+1))dx=((8(√3)π)/(147))](https://www.tinkutara.com/question/Q208777.png)
$$\mathrm{2}\underset{\mathrm{1}/\mathrm{3}} {\overset{\mathrm{1}} {\int}}\frac{{x}\sqrt{−\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{1}}}{\mathrm{7}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{arcsin}\:\left(\mathrm{3}{x}−\mathrm{2}\right)\:\rightarrow\:{dx}=\frac{\sqrt{−\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{1}}}{\:\sqrt{\mathrm{3}}}{dt}\right] \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\:\mathrm{3}}\underset{−\pi/\mathrm{2}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\left(\mathrm{2}+\mathrm{sin}\:{t}\right)\mathrm{cos}^{\mathrm{2}} \:{t}}{\mathrm{7sin}^{\mathrm{2}} \:{t}\:+\mathrm{16sin}\:{t}\:+\mathrm{13}}{dt}= \\ $$$$=\frac{\mathrm{24}\sqrt{\mathrm{3}}}{\mathrm{49}}\underset{−\pi/\mathrm{2}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\mathrm{2}+\mathrm{3sin}\:{t}}{\mathrm{7sin}^{\mathrm{2}} \:{t}\:+\mathrm{16sin}\:{t}\:+\mathrm{13}}{dt}+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{147}}\underset{−\pi/\mathrm{2}} {\overset{\pi/\mathrm{2}} {\int}}\left(\mathrm{2}−\mathrm{7sin}\:{t}\right){dt} \\ $$$$\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{147}}\underset{−\pi/\mathrm{2}} {\overset{\pi/\mathrm{2}} {\int}}\left(\mathrm{2}−\mathrm{7sin}\:{t}\right){dt}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{147}}\left[\mathrm{2}{t}+\mathrm{7cos}\:{t}\right]_{−\pi/\mathrm{2}} ^{\pi/\mathrm{2}} = \\ $$$$=\frac{\mathrm{4}\sqrt{\mathrm{3}}\pi}{\mathrm{147}} \\ $$$$\frac{\mathrm{24}\sqrt{\mathrm{3}}}{\mathrm{49}}\underset{−\pi/\mathrm{2}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\mathrm{2}+\mathrm{3sin}\:{t}}{\mathrm{7sin}^{\mathrm{2}} \:{t}\:+\mathrm{16sin}\:{t}\:+\mathrm{13}}{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:\rightarrow\:{dt}=\frac{\mathrm{2}{du}}{{u}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=\frac{\mathrm{96}\sqrt{\mathrm{3}}}{\mathrm{49}}\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\:\frac{{u}^{\mathrm{2}} +\mathrm{3}{u}+\mathrm{1}}{\mathrm{13}{u}^{\mathrm{4}} +\mathrm{32}{u}^{\mathrm{3}} +\mathrm{54}{u}^{\mathrm{2}} +\mathrm{32}{u}+\mathrm{13}}{du}= \\ $$$$=\frac{\mathrm{96}\sqrt{\mathrm{3}}}{\mathrm{637}}\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\frac{{u}^{\mathrm{2}} +\mathrm{3}{u}+\mathrm{1}}{\Pi\left({u}^{\mathrm{2}} +\frac{\mathrm{4}\left(\mathrm{4}\pm\sqrt{\mathrm{3}}\right)}{\mathrm{13}}{u}+\frac{\mathrm{19}\pm\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{13}}\right)}{du}= \\ $$$$={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{16}}{\mathrm{49}}\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\frac{{u}+\frac{\mathrm{29}−\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{52}}}{{u}^{\mathrm{2}} +\frac{\mathrm{4}\left(\mathrm{4}−\sqrt{\mathrm{3}}\right)}{\mathrm{13}}{u}+\frac{\mathrm{19}−\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{13}}}{du} \\ $$$${I}_{\mathrm{2}} =−\frac{\mathrm{16}}{\mathrm{49}}\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\frac{{u}+\frac{\mathrm{29}+\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{52}}}{{u}^{\mathrm{2}} +\frac{\mathrm{4}\left(\mathrm{4}+\sqrt{\mathrm{3}}\right)}{\mathrm{13}}{u}+\frac{\mathrm{19}+\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{13}}}{du} \\ $$$${I}_{\mathrm{1}} =\left[\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{147}}\mathrm{arctan}\:\frac{\left(\mathrm{4}+\sqrt{\mathrm{3}}\right){u}+\mathrm{2}}{\mathrm{3}}\:+\frac{\mathrm{8}}{\mathrm{49}}\mathrm{ln}\:\left({u}^{\mathrm{2}} +\frac{\mathrm{4}\left(\mathrm{4}−\sqrt{\mathrm{3}}\right)}{\mathrm{13}}{u}+\frac{\mathrm{19}−\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{13}}\right)\right]_{−\mathrm{1}} ^{\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\left[\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{147}}\mathrm{arctan}\:\frac{\left(\mathrm{4}−\sqrt{\mathrm{3}}\right){u}+\mathrm{2}}{\mathrm{3}}\:−\frac{\mathrm{8}}{\mathrm{49}}\mathrm{ln}\:\left({u}^{\mathrm{2}} +\frac{\mathrm{4}\left(\mathrm{4}+\sqrt{\mathrm{3}}\right)}{\mathrm{13}}{u}+\frac{\mathrm{19}+\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{13}}\right)\right]_{−\mathrm{1}} ^{\mathrm{1}} \\ $$$${I}_{\mathrm{1}} +{I}_{\mathrm{2}} =\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{147}}\left(\mathrm{arctan}\:\frac{\mathrm{6}+\sqrt{\mathrm{3}}}{\mathrm{3}}\:+\mathrm{ectan}\:\frac{\mathrm{6}−\sqrt{\mathrm{3}}}{\mathrm{3}}\:+\mathrm{arctan}\:\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{3}}\:+\mathrm{arctan}\:\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{3}}\right)= \\ $$$$=\frac{\mathrm{4}\sqrt{\mathrm{3}}\pi}{\mathrm{147}} \\ $$$$\Rightarrow \\ $$$$\mathrm{2}\underset{\mathrm{1}/\mathrm{3}} {\overset{\mathrm{1}} {\int}}\frac{{x}\sqrt{−\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{1}}}{\mathrm{7}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}}{dx}=\frac{\mathrm{8}\sqrt{\mathrm{3}}\pi}{\mathrm{147}} \\ $$
Commented by Frix last updated on 24/Jun/24
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