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Question Number 154621 by liberty last updated on 20/Sep/21

$$\int \frac{\sqrt{x^2+x+1}}{x^2+x}dx$$

Commented by mathdanisur last updated on 20/Sep/21

$$=\int \underset{}{\underbrace{\frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}}}+\int \underset{}{\underbrace{\frac{\mathrm{dx}}{({\mathrm{x}}^2+\mathrm{x})\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}}}$$$$\left(1\right)\Rightarrow \ln \mid \mathrm{x}+\frac{1}{2}+\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\mid \mathrm{and}\left(2\right)\Rightarrow \ln \mid \frac{{2\mathrm{x}}^2+2\mathrm{x}-1+(2\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{{2\mathrm{x}}^2+2\mathrm{x}+1+(2\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}\mid$$$$\mathrm{say}\to \mathbf{s}=\mathrm{x}+\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\Rightarrow \mathrm{dx}=\frac{2({\mathrm{s}}^2+\mathrm{s}+1)}{{(2\mathrm{s}+1)}^2}\mathrm{ds}$$$$...\Rightarrow =\ln \mid \frac{{2\mathrm{x}}^2+2\mathrm{x}-1+(2\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{{2\mathrm{x}}^2+2\mathrm{x}+1+(2\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}\mid +\mathbb{C}$$

Answered by ARUNG_Brandon_MBU last updated on 20/Sep/21

$$I=\int \frac{\sqrt{x^2+x+1}}{x^2+x}dx=\int \frac{x^2+x+1}{(x^2+x)\sqrt{x^2+x+1}}dx$$$$=\int \frac{dx}{\sqrt{x^2+x+1}}+\int \frac{dx}{(x^2+x)\sqrt{x^2+x+1}}$$$$=\int \frac{dx}{\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}}+\int (\frac{1}{x}-\frac{1}{x+1})\frac{dx}{\sqrt{x^2+x+1}}$$$$=\mathrm{argsinh}\left(\frac{2x+1}{\sqrt{3}}\right)+\int \frac{dx}{x\sqrt{x^2+x+1}}-\int \frac{dx}{(x+1)\sqrt{x^2+x+1}}$$$$=\ln \mid \left(\frac{2x+1}{\sqrt{3}}\right)+\sqrt{1+{\left(\frac{2x+1}{\sqrt{3}}\right)}^2}\mid -\int \frac{udu}{u^2\sqrt{\frac{1}{u^2}+\frac{1}{u}+1}}$$$$+\int \frac{vdv}{v^2\sqrt{{(\frac{1}{v}-1)}^2+(\frac{1}{v}-1)+1}}$$$$\{\begin{array}{l}u=\frac{1}{x}\\ v=\frac{1}{x+1}\end{array}\Rightarrow \{\begin{array}{l}du=-\frac{dx}{x^2}\\ dv=-\frac{dx}{{(x+1)}^2}\end{array}\Rightarrow \{\begin{array}{l}-\frac{1}{u^2}du=dx\\ -\frac{1}{v^2}dv=dx\end{array}$$$$I=\ln \mid \left(\frac{2x+1}{\sqrt{3}}\right)+\sqrt{1+{\left(\frac{2x+1}{\sqrt{3}}\right)}^2}\mid -\mathrm{sgn}\left(u\right)\int \frac{du}{\sqrt{u^2+u+1}}+\mathrm{sgn}\left(v\right)\int \frac{dv}{\sqrt{v^2-v+1}}$$$$=\ln \mid \left(\frac{2x+1}{\sqrt{3}}\right)+\sqrt{1+{\left(\frac{2x+1}{\sqrt{3}}\right)}^2}\mid -\mathrm{sgn}\left(u\right)\ln \mid \left(\frac{2u+1}{\sqrt{3}}\right)+\sqrt{1+{\left(\frac{2u+1}{\sqrt{3}}\right)}^2}\mid$$$$+\mathrm{sgn}\left(v\right)\ln \mid \left(\frac{2v-1}{\sqrt{3}}\right)+\sqrt{1+{\left(\frac{2v-1}{\sqrt{3}}\right)}^2}\mid +C$$

Answered by EDWIN88 last updated on 20/Sep/21

$$I=\int \frac{\sqrt{x^2+x+1}+1-1}{(x^2+x+1)-1}dx$$$$I=\int \frac{\sqrt{x^2+x+1}-1}{(\sqrt{x^2+x+1}-1).(\sqrt{x^2+x+1}+1)}dx$$$$+\int \frac{dx}{x^2+x}$$$$I_1=\int \frac{dx}{\sqrt{x^2+x+1}+1}=\int \frac{dx}{\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}+1}$$$$setx+\frac{1}{2}=\frac{\sqrt{3}}{2}\tan \alpha$$$$I_1=\int \frac{\sec {}^2\alpha (\sec \alpha -1)}{(\sec \alpha +1)(\sec \alpha -1)}d\alpha$$$$I_1=\int \frac{\sec {}^3\alpha -\sec {}^2\alpha }{\tan {}^2\alpha }d\alpha$$$$I_1=\ln (\sec \left({\tan }^{-1}(\frac{2}{\sqrt{3}}x+\frac{1}{\sqrt{3}})\right)+\frac{2}{\sqrt{3}}x+\frac{1}{\sqrt{3}})-$$$$\frac{1}{\sin \left({\tan }^{-1}(\frac{2}{\sqrt{3}}x+\frac{1}{\sqrt{3}})\right)}-\frac{1}{(\frac{2}{\sqrt{3}}x+\frac{1}{\sqrt{3}})}+c_1$$$$I_2=\int \frac{dx}{x^2+x}=\int \frac{dx}{x(x+1)}=\int (\frac{1}{x}-\frac{1}{x+1})dx$$$$I_2=\ln x-\ln (x+1)+c_2$$$$\therefore I=I_1+I_2$$

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