2-1-3-1-x-3x-2-4x-1-7x-2-4x-1-dx-Exact-solution-needed-

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

2 \underset{\frac{1}{3}}{\int^{1}} \frac{x \sqrt{- 3 x^{2} + 4 x - 1}}{7 x^{2} - 4 x + 1} d x = ?

Exact solution needed .

Answered by Ghisom last updated on 24/Jun/24

2 \underset{1 / 3}{\int^{1}} \frac{x \sqrt{- 3 x^{2} + 4 x - 1}}{7 x^{2} - 4 x + 1} d x =

[t = \arcsin (3 x - 2) \to d x = \frac{\sqrt{- 3 x^{2} + 4 x - 1}}{\sqrt{3}} d t]

= \frac{2 \sqrt{3}}{3} \underset{- π / 2}{\int^{π / 2}} \frac{(2 + \sin t) \cos^{2} t}{{7 \sin}^{2} t + 16 \sin t + 13} d t =

= \frac{24 \sqrt{3}}{49} \underset{- π / 2}{\int^{π / 2}} \frac{2 + 3 \sin t}{{7 \sin}^{2} t + 16 \sin t + 13} d t + \frac{2 \sqrt{3}}{147} \underset{- π / 2}{\int^{π / 2}} (2 - 7 \sin t) d t

\frac{2 \sqrt{3}}{147} \underset{- π / 2}{\int^{π / 2}} (2 - 7 \sin t) d t = \frac{2 \sqrt{3}}{147} {[2 t + 7 \cos t]}_{- π / 2}^{π / 2} =

= \frac{4 \sqrt{3} π}{147}

\frac{24 \sqrt{3}}{49} \underset{- π / 2}{\int^{π / 2}} \frac{2 + 3 \sin t}{{7 \sin}^{2} t + 16 \sin t + 13} d t =

[u = \tan \frac{t}{2} \to d t = \frac{2 d u}{u^{2} + 1}]

= \frac{96 \sqrt{3}}{49} \underset{- 1}{\int^{1}} \frac{u^{2} + 3 u + 1}{13 u^{4} + 32 u^{3} + 54 u^{2} + 32 u + 13} d u =

= \frac{96 \sqrt{3}}{637} \underset{- 1}{\int^{1}} \frac{u^{2} + 3 u + 1}{Π (u^{2} + \frac{4 (4 \pm \sqrt{3})}{13} u + \frac{19 \pm 8 \sqrt{3}}{13})} d u =

= I_{1} + I_{2}

I_{1} = \frac{16}{49} \underset{- 1}{\int^{1}} \frac{u + \frac{29 - 4 \sqrt{3}}{52}}{u^{2} + \frac{4 (4 - \sqrt{3})}{13} u + \frac{19 - 8 \sqrt{3}}{13}} d u

I_{2} = - \frac{16}{49} \underset{- 1}{\int^{1}} \frac{u + \frac{29 + 4 \sqrt{3}}{52}}{u^{2} + \frac{4 (4 + \sqrt{3})}{13} u + \frac{19 + 8 \sqrt{3}}{13}} d u

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} + ectan \frac{6 - \sqrt{3}}{3} + \arctan \frac{2 + \sqrt{3}}{3} + \arctan \frac{2 - \sqrt{3}}{3}) =

= \frac{4 \sqrt{3} π}{147}

\Rightarrow

2 \underset{1 / 3}{\int^{1}} \frac{x \sqrt{- 3 x^{2} + 4 x - 1}}{7 x^{2} - 4 x + 1} d x = \frac{8 \sqrt{3} π}{147}

Commented by Frix last updated on 24/Jun/24

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