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Question Number 124261 by Ar Brandon last updated on 02/Dec/20

$$\int \frac{{2\mathrm{x}}^2+5\mathrm{x}+9}{(\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}\mathrm{dx}$$

Commented by MJS_new last updated on 02/Dec/20

$$\mathrm{I}\mathrm{think}\mathrm{something}\mathrm{went}\mathrm{wrong}...$$

Commented by Ar Brandon last updated on 02/Dec/20

$$\mathrm{I}\mathrm{get}$$$$\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\mathrm{x}+1}{\sqrt{3}}\right)$$$$-6\ln \mid \frac{1-\mathrm{x}+2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{2(\mathrm{x}+1)}\mid +\mathrm{K}$$$$Anyapproval,please?$$

Answered by MJS_new last updated on 02/Dec/20

$$\int \frac{2x^2+5x+9}{(x+1)\sqrt{x^2+x+1}}dx=$$$$[t=x+\frac{1}{2}\to dx=dt]$$$$=4\int \frac{2t^2+3t+7}{(2t+1)\sqrt{4t^2+3}}dt=$$$$[u=\frac{\sqrt{3}(2t+\sqrt{4t^2+3})}{3}\to dt=\frac{\sqrt{3(4t^2+3)}}{2(2t+\sqrt{4t^2+3})}]$$$$=\int \frac{3\sqrt{3}{u}^4+18u^3+50\sqrt{3}{u}^2-18u+3\sqrt{3}}{2u^2(3u^2+2\sqrt{3}u-3)}du=$$$$=\int (\frac{\sqrt{3}}{2}-\frac{6}{u+\sqrt{3}}+\frac{2}{u}+\frac{6}{u-\frac{\sqrt{3}}{3}}-\frac{\sqrt{3}}{2u^2})du=$$$$=\frac{\sqrt{3}}{2}u-6\ln (u+\sqrt{3})+2\ln u+6\ln (u-\frac{\sqrt{3}}{3})+\frac{\sqrt{3}}{2u}=$$$$=\frac{\sqrt{3}(u^2+1)}{2u}+6\ln \frac{3u-\sqrt{3}}{u+\sqrt{3}}+2\ln u=$$$$...$$$$=2\sqrt{x^2+x+1}+6\ln \frac{x-1+2\sqrt{x^2+x+1}}{x+1}+2\ln (2x+1+2\sqrt{x^2+x+1})+C$$

Commented by Ar Brandon last updated on 02/Dec/20

Thanks Sir, I found my mistake.

Commented by Ar Brandon last updated on 02/Dec/20

��Lol, OK ��

Commented by Ar Brandon last updated on 02/Dec/20

By the way. I left a message for you privately. Can you please check ?��

Commented by Ar Brandon last updated on 02/Dec/20

I don't think so, bro. Perhaps you're talking about the difference in sign. It's due to the absolute value of "u" I think.

Commented by Ar Brandon last updated on 02/Dec/20

OK. Thanks for your opinion, Sir. I'll check and post my solution.

Commented by Ar Brandon last updated on 02/Dec/20

Sir, I don't understand how you decide on your change of variables everytime. And they sound pretty fast��

Commented by MJS_new last updated on 02/Dec/20

$$\mathrm{I}\mathrm{hate}\mathrm{to}\mathrm{deal}\mathrm{with}\sinh \mathrm{and}\cosh \mathrm{so}\mathrm{I}\mathrm{usually}$$$$\mathrm{use}\mathrm{these}:$$$$\sqrt{{ax}^2\pm b}\Rightarrow t=\frac{\sqrt{a}x+\sqrt{{ax}^2\pm b}}{\sqrt{b}}\iff x=\frac{\sqrt{b}(t^2\mp 1)}{2\sqrt{a}t}\to dx=\frac{\sqrt{b({ax}^2\pm b)}}{ax+\sqrt{a({ax}^2\pm b)}}$$

Commented by Ar Brandon last updated on 02/Dec/20

Oh, I see ! ��

Commented by Dwaipayan Shikari last updated on 02/Dec/20

$$Oneortwomoresir:)$$

Commented by Dwaipayan Shikari last updated on 02/Dec/20

I have told that sir may share one or more examples regarding this process ��

Commented by Ar Brandon last updated on 02/Dec/20

No, I meant on telegram. Sorry I didn't precise.

Answered by Ar Brandon last updated on 02/Dec/20

$$\mathcal{I}=\int \frac{{2\mathrm{x}}^2+5\mathrm{x}+9}{(\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}\mathrm{dx}$$$$\frac{{2\mathrm{x}}^2+5\mathrm{x}+9}{\mathrm{x}+1}=\frac{2{(\mathrm{x}+1)}^2+(\mathrm{x}+1)+6}{\mathrm{x}+1}=2(\mathrm{x}+1)+1+\frac{6}{\mathrm{x}+1}$$$$\mathcal{I}=\int \{2(\mathrm{x}+1)+1+\frac{6}{\mathrm{x}+1}\}\cdot \frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}$$$$=2\int \frac{\mathrm{x}+1}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}\mathrm{dx}+\int \frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}+6\int \frac{\mathrm{dx}}{(\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}$$$$=\int \frac{2\mathrm{x}+1}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}\mathrm{dx}+2\int \frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}+6\int \frac{\mathrm{dx}}{(\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}$$$$=2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}+2\ln \mid (\mathrm{x}+\frac{1}{2})+\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\mid +6\mathcal{J}$$$$\mathrm{For}\mathcal{J},$$$$\mathrm{Let}\frac{1}{\mathrm{x}+1}=\mathrm{u}\Rightarrow -\frac{\mathrm{dx}}{{(\mathrm{x}+1)}^2}=\mathrm{du}\Rightarrow \mathrm{dx}=-\frac{\mathrm{du}}{{\mathrm{u}}^2}$$$$\Rightarrow \mathrm{x}=\frac{1}{\mathrm{u}}-1,{\mathrm{x}}^2+\mathrm{x}+1=\frac{1}{{\mathrm{u}}^2}-\frac{1}{\mathrm{u}}+1$$$$\mathcal{J}=-\int \frac{\mathrm{u}}{\sqrt{\frac{1}{{\mathrm{u}}^2}-\frac{1}{\mathrm{u}}+1}}\cdot \frac{\mathrm{du}}{{\mathrm{u}}^2}=-\int \frac{\mathrm{du}}{\sqrt{1-\mathrm{u}+{\mathrm{u}}^2}}$$$$=-\int \frac{\mathrm{du}}{\sqrt{{(\mathrm{u}-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}}=-{\sinh }^{-1}\left(\frac{\mathrm{u}-1/2}{\sqrt{3}/2}\right)+\mathrm{C}$$$$=-\ln \mid (\mathrm{u}-\frac{1}{2})+\sqrt{{(\mathrm{u}-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\mid +\mathrm{C}$$$$=-\ln \mid (\frac{1}{\mathrm{x}+1}-\frac{1}{2})+\sqrt{\frac{1}{{(\mathrm{x}+1)}^2}-\frac{1}{\mathrm{x}+1}+1}\mid +\mathrm{C}$$$$=-\ln \mid \frac{1-\mathrm{x}+2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{2(\mathrm{x}+1)}\mid +\mathrm{C}$$$$\Rightarrow \mathcal{I}=2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}+2\ln \mid (\mathrm{x}+\frac{1}{2})+\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\mid -6\ln \mid \frac{1-\mathrm{x}+2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{2(\mathrm{x}+1)}\mid +\mathrm{K}$$

Answered by Dwaipayan Shikari last updated on 02/Dec/20

$$=\int \frac{(2x+1)}{\sqrt{x^2+x+1}}dx+\int \frac{2x+8}{(x+1)\sqrt{x^2+x+1}}dx$$$$=2\sqrt{x^2+x+1}+2\int \frac{1}{\sqrt{x^2+x+1}}+6\int \frac{1}{(x+1)\sqrt{x^2+x+1}}dx(x+1)=t$$$$=2\sqrt{x^2+x+1}+2\int \frac{1}{\sqrt{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}}+6\int \frac{1}{t\sqrt{t^2-t+1}}dt,t=\frac{1}{u}$$$$=2\sqrt{x^2+x+1}+2log(x+\frac{1}{2}+\sqrt{x^2+x+1})-6\int \frac{1}{\sqrt{u^2-u+1}}du$$$$=2(\sqrt{x^2+x+1}+log(x+\frac{1}{2}+\sqrt{x^2+x+1}))-6log(u-\frac{1}{2}+\sqrt{u^2-u+1})$$$$2\sqrt{x^2+x+1}+log(x+\frac{1}{2}+\sqrt{x^2+x+1})-6log(\frac{1}{x+1}-\frac{1}{2}+\sqrt{\frac{1}{{(x+1)}^2}-\frac{1}{(x+1)}-1})+C$$

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