4x-2-3-x-2-x-1-2-dx-

Question Number 52649 by Tawa1 last updated on 10/Jan/19

\int \frac{{4 x}^{2} + 3}{{(x^{2} + x + 1)}^{2}} dx

Commented by maxmathsup by imad last updated on 10/Jan/19

e t I = \int \frac{4 x^{2} + 3}{{(x^{2} + x + 1)}^{2}} d x \Rightarrow I = \int \frac{4 (x^{2} + x + 1) - 4 x - 1}{{(x^{2} + x + 1)}^{2}} d x

= 4 \int \frac{d x}{x^{2} + x + 1} - \int \frac{4 x + 1}{{(x^{2} + x + 1)}^{2}} d x = 4 H - K

H = \int \frac{d x}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}} =_{x + \frac{1}{2} = \frac{\sqrt{3}}{2} t a n θ} \int \frac{1}{\frac{3}{4} (1 + {t a n}^{2} θ)} \frac{\sqrt{3}}{2} (1 + {t a n}^{2} θ) d θ

= \frac{\sqrt{3}}{2} \frac{4}{3} \int d θ = \frac{2}{\sqrt{3}} θ + c_{1} = \frac{2}{\sqrt{3}} a r c t a n (\frac{2 x + 1}{\sqrt{3}}) + c_{1}

w e h a v e {(\frac{1}{x^{2} + x + 1})}^{^{'}} = - \frac{2 x + 1}{(x^{2} + x + 1)} \Rightarrow

\int \frac{4 x + 1}{{(x^{2} + x + 1)}^{2}} = 2 \int \frac{2 x + \frac{1}{2}}{{(x^{2} + x + 1)}^{2}} d x = 2 \int \frac{2 x + 1 - \frac{1}{2}}{{(x^{2} + x + 1)}^{2}} d x

= 2 {- \frac{1}{x^{2} + x + 1} - \frac{1}{2} \int \frac{d x}{{(x^{2} + x + 1)}^{2}}}

= \frac{- 2}{x^{2} + x + 1} - \int \frac{d x}{{(x^{2} + x + 1)}^{2}} b u t

\int \frac{d x}{{(x^{2} + x + 1)}^{2}} = \int \frac{d x}{{({(x + \frac{1}{2})}^{2} + \frac{3}{4})}^{2}} =_{x + \frac{1}{2} = \frac{\sqrt{3}}{2} t a n θ} \frac{16}{9} \frac{\sqrt{3}}{2} \int \frac{1}{{(1 + {t a n}^{2} θ)}^{2}} (1 + {t a n}^{2} θ) d θ

= \frac{8 \sqrt{3}}{9} \int \frac{d θ}{1 + {t a n}^{2} θ} = \frac{8 \sqrt{3}}{9} \int \frac{1 + c o s (2 θ)}{2} d θ = \frac{4 \sqrt{3}}{9} \int (1 + c o s (2 θ)) d θ

= \frac{4 \sqrt{3}}{9} θ + \frac{4 \sqrt{3}}{18} s i n (2 θ) = \frac{4 \sqrt{3}}{9} a r c t a n (\frac{2 x + 1}{\sqrt{3}}) + \frac{2 \sqrt{3}}{9} s i n (2 a r c t a n (\frac{2 x + 1}{\sqrt{3}})) + c_{2} \Rightarrow

I = \frac{8}{\sqrt{3}} a r c t a n (\frac{2 x + 1}{\sqrt{3}}) + \frac{2}{x^{2} + x + 1} + \frac{4 \sqrt{3}}{9} a r c t a n (\frac{2 x + 1}{\sqrt{3}}) + \frac{2 \sqrt{3}}{9} s i n (2 a r c t a n (\frac{2 x + 1}{\sqrt{3}})) + C

Commented by Tawa1 last updated on 10/Jan/19

God bless you sir

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