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Question Number 113278 by mathdave last updated on 12/Sep/20

$$solve$$$$\int \frac{\sqrt{x^2+x+2-\sqrt{4x^2+4x+4}}}{x\sqrt{x^4+x^3+x^2}}dx$$

Answered by 1549442205PVT last updated on 13/Sep/20

$${\mathrm{x}}^2+\mathrm{x}+2-\sqrt{{4\mathrm{x}}^2+4\mathrm{x}+4}=$$$${\left(\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\right)}^2-2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}+1$$$$={(\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-1)}^2,\mathrm{x}\sqrt{{\mathrm{x}}^4+{\mathrm{x}}^3+{\mathrm{x}}^2}={\mathrm{x}}^2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}$$$$\mathrm{Hence},\mathrm{F}=\int \frac{\sqrt{x^2+x+2-\sqrt{4x^2+4x+4}}}{x\sqrt{x^4+x^3+x^2}}dx$$$$=\int \frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-1}{{\mathrm{x}}^2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}\mathrm{dx}=\int (\frac{1}{{\mathrm{x}}^2}-\frac{1}{{\mathrm{x}}^2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}})\mathrm{dx}$$$$\mathrm{We}\mathrm{find}\mathrm{J}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}=\int \frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{{\mathrm{x}}^2({\mathrm{x}}^2+\mathrm{x}+1)}\mathrm{dx}$$$$=\int \sqrt{{\mathrm{x}}^2+\mathrm{x}+1}(\frac{-\mathrm{x}+1}{{\mathrm{x}}^2}+\frac{\mathrm{x}}{{\mathrm{x}}^2+\mathrm{x}+1})\mathrm{dx}$$$$=\int (\frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{{\mathrm{x}}^2}-\frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{\mathrm{x}}+\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}})\mathrm{dx}$$$$==\int [\frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{{\mathrm{x}}^2}-\frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{\mathrm{x}}-\frac{1}{2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}+\left(\frac{1}{2}\times \frac{2\mathrm{x}+1}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}\right)]\mathrm{dx}$$$$\frac{1}{2}\int \frac{2\mathrm{x}+1}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}\mathrm{dx}=\frac{1}{2}\int \frac{\mathrm{du}}{\sqrt{\mathrm{u}}}(\mathrm{u}={\mathrm{x}}^2+\mathrm{x}+1)$$$$=\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}$$$$\mathrm{Since}\int \frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^2+\lambda }}=\ln \mid \mathrm{x}+\sqrt{{\mathrm{x}}^2+\lambda }\mid +\mathrm{C}$$$$\frac{1}{2}\int \frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}=\frac{1}{2}\int \frac{\mathrm{d}(\mathrm{x}+\frac{1}{2})}{\sqrt{{(\mathrm{x}+\frac{1}{2})}^2+\frac{3}{4}}}=\frac{1}{2}\ln \mid \mathrm{x}+\frac{1}{2}+\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\mid$$$$\mathrm{Therefore},\mathrm{we}\mathrm{just}\mathrm{need}\mathrm{find}$$$$=\int (\frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{{\mathrm{x}}^2}-\frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{\mathrm{x}})]\mathrm{dx}=\mathrm{A}+\mathrm{B}$$$$\mathrm{A}=\int \frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{{\mathrm{x}}^2}=-\int \sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\mathrm{d}\left(\frac{1}{\mathrm{x}}\right)$$$$\underset{\mathrm{by}\mathrm{part}}{=}\frac{-\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{\mathrm{x}}+\int \frac{1}{\mathrm{x}}\times \frac{2\mathrm{x}+1}{2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}\mathrm{dx}$$$$=\frac{-\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{\mathrm{x}}+\int \frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}+\int \frac{\mathrm{dx}}{2\mathrm{x}\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}$$$$=\frac{-\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{\mathrm{x}}+\ln \mid \mathrm{x}+\frac{1}{2}+\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\mid +\frac{1}{2}\mathrm{M}\left(1\right)$$$$\mathrm{B}=\int \frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{\mathrm{x}}\mathrm{dx}\underset{\mathrm{by}\mathrm{parts}}{=}\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-$$$$-\int [\mathrm{x}.\frac{.(\frac{(2\mathrm{x}+1)\mathrm{x}}{2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}-\sqrt{{\mathrm{x}}^2+\mathrm{x}+1})}{{\mathrm{x}}^2}]\mathrm{dx}$$$$=\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\int \frac{-\mathrm{x}-2}{2\mathrm{x}\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}\mathrm{dx}$$$$=\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}+\int \frac{\mathrm{dx}}{2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}+\int \frac{\mathrm{dx}}{\mathrm{x}\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}$$$$=\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}+\frac{1}{2}\ln \mid \mathrm{x}+\frac{1}{2}+\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\mid +\mathrm{M}\left(2\right)$$$$\mathrm{M}=\int \frac{\mathrm{dx}}{\mathrm{x}\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}=\int \left(\frac{2}{2\mathrm{x}\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}\times \frac{2\mathrm{x}+1-2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{2\mathrm{x}+1-2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}\right)\mathrm{dx}$$$$=\int \left(\frac{2\mathrm{x}+1-2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}\times \frac{2}{\mathrm{x}(2\mathrm{x}+1)-2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}\right)\mathrm{dx}$$$$=\int (\frac{2\mathrm{x}+1}{2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}-1)(\frac{1}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\mathrm{x}-1}-\frac{1}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\mathrm{x}+1})\mathrm{dx}$$$$=\int (\frac{\frac{2\mathrm{x}+1}{2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}-1}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\mathrm{x}-1}-\frac{\frac{2\mathrm{x}+1}{2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}-1}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\mathrm{x}+1})\mathrm{dx}$$$$=\ln \mid \sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\mathrm{x}-1\mid -\ln \mid \sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\mathrm{x}+1\mid$$$$=\ln \mid \frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\mathrm{x}-1}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\mathrm{x}+1}\mid \left(3\right)$$$$\mathrm{From}\left(1\right)\left(2\right)\left(3\right)\mathrm{we}\mathrm{get}$$$$\mathrm{A}+\mathrm{B}=-\frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}}{\mathrm{x}}+\ln \mid \mathrm{x}+\frac{1}{2}+\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\mid$$$$+\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}+\frac{1}{2}\ln \mid \mathrm{x}+\frac{1}{2}+\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\mid$$$$+\frac{3}{2}\ln \mid \frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\mathrm{x}-1}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\mathrm{x}+1}\mid \left(4\right)$$$$\mathrm{Therefore},\mathrm{J}=\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\frac{1}{2}\ln \mid \mathrm{x}+\frac{1}{2}+\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\mid$$$$+\mathrm{A}+\mathrm{B}=2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}+\ln \mid \mathrm{x}+\frac{1}{2}+\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\mid +$$$$+\frac{3}{2}\ln \mid \frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\mathrm{x}-1}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\mathrm{x}+1}\mid .\mathrm{From}\mathrm{that}$$$$\mathrm{F}=-\frac{1}{\mathrm{x}}+\mathrm{J}=-\frac{1}{\mathrm{x}}+\mathrm{A}+\mathrm{B}=2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}+\ln \mid \mathrm{x}+\frac{1}{2}+\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\mid +$$$$+\frac{3}{2}\ln \mid \frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\mathrm{x}-1}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\mathrm{x}+1}\mid +\mathrm{C}(\mathrm{C}-\mathrm{constant})$$$$\mathrm{Thus},\mathrm{final}\mathrm{result}\mathrm{is}$$$$\mathrm{F}=\int \frac{\sqrt{x^2+x+2-\sqrt{4x^2+4x+4}}}{x\sqrt{x^4+x^3+x^2}}dx$$$$=-\frac{1}{\mathrm{x}}+2\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}+\ln \mid \mathrm{x}+\frac{1}{2}+\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\mid$$$$+\frac{3}{2}\ln \mid \frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\mathrm{x}-1}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\mathrm{x}+1}\mid +\mathrm{C}$$

Commented by mathdave last updated on 12/Sep/20

$$niceone$$

Commented by 1549442205PVT last updated on 13/Sep/20

$$\mathrm{I}\mathrm{like}\mathrm{smart}{\mathrm{Sir}}^{\prime }\mathrm{s}\mathrm{solution}$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{grest}\mathrm{sir}$$

Answered by MJS_new last updated on 12/Sep/20

$$\int \frac{\sqrt{x^2+x+1-2\sqrt{x^2+x+1}+1}}{x^2\sqrt{x^2+x+1}}dx=$$$$=\int \frac{dx}{x^2}-\int \frac{dx}{x^2\sqrt{x^2+x+1}}$$$$\left(1\right)\int \frac{dx}{x^2}=-\frac{1}{x}$$$$\left(2\right)-\int \frac{dx}{x^2\sqrt{x^2+x+1}}=$$$$[t=\frac{\sqrt{3}}{2}(2x+1+2\sqrt{x^2+x+1}\to dx=\frac{\sqrt{3(x^2+x+1)}}{2x+1+2\sqrt{x^2+x+1}}]$$$$=-16\int \frac{t}{{(\sqrt{3}{t}^2-2t-\sqrt{3})}^2}dt=$$$$=\sqrt{3}\int (\frac{1}{2(\sqrt{3}t-3)}-\frac{1}{2(\sqrt{3}t+1)}-\frac{3}{{(\sqrt{3}t-3)}^2}+\frac{1}{{(\sqrt{3}t+2)}^2})dt=$$$$=\frac{2(t+\sqrt{3})}{\sqrt{3}{t}^2-2t-\sqrt{3}}-\frac{1}{2}\ln \frac{\sqrt{3}t+1}{\sqrt{3}t-3}=$$$$=\frac{\sqrt{x^2+x+1}}{x}+\frac{1}{2}\ln x-\frac{1}{2}\ln (x+2+2\sqrt{x^2+x+1})+C$$$$$$$$\Rightarrow$$$$$$$$\int \frac{\sqrt{x^2+x+2-\sqrt{4x^2+4x+4}}}{x\sqrt{x^4+x^3+x^2}}dx=$$$$=\frac{-1+\sqrt{x^2+x+1}}{x}+\frac{1}{2}\ln x-\frac{1}{2}\ln (x+2+2\sqrt{x^2+x+1})+C$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}$$

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