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Question Number 193982 by cortano12 last updated on 25/Jun/23

$$\int \frac{{6\mathrm{x}}^3+{9\mathrm{x}}^2+15\mathrm{x}+6}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}\mathrm{dx}=?$$

Answered by MM42 last updated on 26/Jun/23

$$I=({ax}^2+bx+e)\sqrt{x^2+x+1}+f\int \frac{dx}{\sqrt{x^2+c+1}}$$$$\stackrel{d}{\Rightarrow }\frac{6x^3+9x^2+15x+6}{\sqrt{x^2+x+1}}=(2ax+b)\sqrt{x^2+x+1}+\frac{({ax}^2+bx+e)(2x+1)}{2\sqrt{x^2+x+1}}+\frac{f}{\sqrt{x^2+x+1}}$$$$\Rightarrow a=2,b=\frac{4}{3},e=\frac{25}{3},f=-\frac{35}{3}$$$$I=(2x^2+\frac{4}{3}x+\frac{25}{3})\sqrt{x^2+x+1}-\frac{35}{3}\int \frac{dx}{\sqrt{x^2+x+1}}$$$$★\int \frac{dx}{\sqrt{x^2+x+1}}=\int \frac{dx}{\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}}=ln(x+\frac{1}{2}+\sqrt{x^2+x+1})+c$$$$$$

Commented by Frix last updated on 25/Jun/23

$$?$$

Answered by Frix last updated on 25/Jun/23

$$t=\frac{2x+1+2\sqrt{x^2+x+1}}{\sqrt{3}}\Rightarrow dx=\frac{\sqrt{x^2+x+1}}{t}dt$$$$\int \frac{6x^3+9x^2+15x+6}{\sqrt{x^2+x+1}}dx=$$$$=\int (\frac{9\sqrt{3}{t}^2}{32}+\frac{57\sqrt{3}}{32}-\frac{57\sqrt{3}}{32t^2}-\frac{9\sqrt{3}}{32t^4})dt=$$$$=\frac{3\sqrt{3}{t}^3}{32}+\frac{57\sqrt{3}t}{32}+\frac{57\sqrt{3}}{32t}+\frac{3\sqrt{3}}{32t^3}=$$$$...$$$$=2(x^2+x+4)\sqrt{x^2+x+1}+C$$

Answered by horsebrand11 last updated on 26/Jun/23

$$\mathcal{I}\underset{―}{\mathbf{Z}}\underbrace{}$$

Commented by Frix last updated on 26/Jun/23

$$\mathrm{Test}\mathrm{your}\mathrm{solution}$$$$\frac{d\left[2\sqrt{x^2+x+1}(x^2+x+2)\right]}{dx}=\frac{6x^3+9x^2+11x+4}{\sqrt{x^2+x+1}}$$

Commented by Frix last updated on 26/Jun/23

$$\mathrm{Now}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{correct}.$$

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