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Question Number 208733 by Frix last updated on 22/Jun/24

$$2\underset{\frac{1}{3}}{\stackrel{1}{\int }}\frac{x\sqrt{-3x^2+4x-1}}{7x^2-4x+1}dx=?$$$$\mathrm{Exact}\mathrm{solution}\mathrm{needed}.$$

Answered by Ghisom last updated on 24/Jun/24

$$2\underset{1/3}{\stackrel{1}{\int }}\frac{x\sqrt{-3x^2+4x-1}}{7x^2-4x+1}dx=$$$$[t=\arcsin (3x-2)\to dx=\frac{\sqrt{-3x^2+4x-1}}{\sqrt{3}}dt]$$$$=\frac{2\sqrt{3}}{3}\underset{-\pi /2}{\stackrel{\pi /2}{\int }}\frac{(2+\sin t){\cos }^{2}t}{{7\sin }^{2}t+16\sin t+13}dt=$$$$=\frac{24\sqrt{3}}{49}\underset{-\pi /2}{\stackrel{\pi /2}{\int }}\frac{2+3\sin t}{{7\sin }^{2}t+16\sin t+13}dt+\frac{2\sqrt{3}}{147}\underset{-\pi /2}{\stackrel{\pi /2}{\int }}(2-7\sin t)dt$$$$\frac{2\sqrt{3}}{147}\underset{-\pi /2}{\stackrel{\pi /2}{\int }}(2-7\sin t)dt=\frac{2\sqrt{3}}{147}{[2t+7\cos t]}_{-\pi /2}^{\pi /2}=$$$$=\frac{4\sqrt{3}\pi }{147}$$$$\frac{24\sqrt{3}}{49}\underset{-\pi /2}{\stackrel{\pi /2}{\int }}\frac{2+3\sin t}{{7\sin }^{2}t+16\sin t+13}dt=$$$$[u=\tan \frac{t}{2}\to dt=\frac{2du}{u^2+1}]$$$$=\frac{96\sqrt{3}}{49}\underset{-1}{\stackrel{1}{\int }}\frac{u^2+3u+1}{13u^4+32u^3+54u^2+32u+13}du=$$$$=\frac{96\sqrt{3}}{637}\underset{-1}{\stackrel{1}{\int }}\frac{u^2+3u+1}{\mathrm{\Pi }(u^2+\frac{4(4\pm \sqrt{3})}{13}u+\frac{19\pm 8\sqrt{3}}{13})}du=$$$$=I_1+I_2$$$$I_1=\frac{16}{49}\underset{-1}{\stackrel{1}{\int }}\frac{u+\frac{29-4\sqrt{3}}{52}}{u^2+\frac{4(4-\sqrt{3})}{13}u+\frac{19-8\sqrt{3}}{13}}du$$$$I_2=-\frac{16}{49}\underset{-1}{\stackrel{1}{\int }}\frac{u+\frac{29+4\sqrt{3}}{52}}{u^2+\frac{4(4+\sqrt{3})}{13}u+\frac{19+8\sqrt{3}}{13}}du$$$$I_1={[\frac{4\sqrt{3}}{147}\arctan \frac{(4+\sqrt{3})u+2}{3}+\frac{8}{49}\ln (u^2+\frac{4(4-\sqrt{3})}{13}u+\frac{19-8\sqrt{3}}{13})]}_{-1}^1$$$$I_2={[\frac{4\sqrt{3}}{147}\arctan \frac{(4-\sqrt{3})u+2}{3}-\frac{8}{49}\ln (u^2+\frac{4(4+\sqrt{3})}{13}u+\frac{19+8\sqrt{3}}{13})]}_{-1}^1$$$$I_1+I_2=\frac{4\sqrt{3}}{147}(\arctan \frac{6+\sqrt{3}}{3}+\mathrm{ectan}\frac{6-\sqrt{3}}{3}+\arctan \frac{2+\sqrt{3}}{3}+\arctan \frac{2-\sqrt{3}}{3})=$$$$=\frac{4\sqrt{3}\pi }{147}$$$$\Rightarrow$$$$2\underset{1/3}{\stackrel{1}{\int }}\frac{x\sqrt{-3x^2+4x-1}}{7x^2-4x+1}dx=\frac{8\sqrt{3}\pi }{147}$$

Commented by Frix last updated on 24/Jun/24

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