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Question Number 158775 by HongKing last updated on 08/Nov/21

$$\mathrm{Find}:$$ $$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{1}{({\mathrm{x}}^2+\mathrm{x}+1)\cdot (1+\mathrm{ax})}\mathrm{dx};\mathrm{a}>0$$ $$$$ $$\mathrm{Answer}:$$ $$\frac{-\pi \sqrt{3}\cdot (\mathrm{a}-2)+9\mathrm{a}\cdot \ln \left(\mathrm{a}\right)}{9\cdot ({\mathrm{a}}^2-\mathrm{a}+1)}$$

Answered by MJS_new last updated on 08/Nov/21

$$\int \frac{dx}{(x^2+x+1)(ax+1)}=$$ $$=\frac{a^2}{a^2-a+1}\int \frac{dx}{ax+1}-\frac{a}{a^2-a+1}\int \frac{x+\frac{a-1}{a}}{x^2+x+1}dx=$$ $$=\frac{a}{a^2-a+1}\ln \mid ax+1\mid -\frac{(a-2)}{\sqrt{3}(a^2-a+1)}\arctan \frac{2x+1}{\sqrt{3}}-\frac{a}{2(a^2-a+1)}\ln (x^2+x+1)=$$ $$=\frac{a}{2(a^2-a+1)}\ln \frac{{(ax+1)}^2}{x^2+x+1}-\frac{(a-2)}{\sqrt{3}(a^2-a+1)}\arctan \frac{2x+1}{\sqrt{3}}+C$$ $$\Rightarrow$$ $$\mathrm{answer}\mathrm{is}$$ $$\frac{9a\ln a-(a-2)\pi \sqrt{3}}{9(a^2-a+1)}$$

Commented byHongKing last updated on 09/Nov/21

$$\mathrm{peefet}\mathrm{my}\mathrm{dear}\mathbf{S}\mathrm{er},\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

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