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Question Number 129014 by bramlexs22 last updated on 12/Jan/21

$${\int }_{-\frac{1}{2}}^{-\frac{1}{4}}x(x+1)\sqrt{1+\frac{1}{x^2}+\frac{1}{{(x+1)}^2}}dx=?$$

Commented by Ajao yinka last updated on 12/Jan/21

37/192

Answered by liberty last updated on 12/Jan/21

$$\mathrm{x}(\mathrm{x}+1)\sqrt{\frac{{\mathrm{x}}^2{(\mathrm{x}+1)}^2+{(\mathrm{x}+1)}^2+{\mathrm{x}}^2}{{\mathrm{x}}^2{(\mathrm{x}+1)}^2}}=$$$$\frac{\sqrt{{\mathrm{x}}^2({\mathrm{x}}^2+2\mathrm{x}+1)+{\mathrm{x}}^2+2\mathrm{x}+1+{\mathrm{x}}^2}}{-1}\mathrm{since}$$$$\sqrt{{\mathrm{x}}^2{(\mathrm{x}+1)}^2}=\mid \mathrm{x}(\mathrm{x}+1)\mid =-\mathrm{x}(\mathrm{x}+1)\mathrm{for}-\frac{1}{4}⩽\mathrm{x}⩽-\frac{1}{2}$$$$\sqrt{{\mathrm{x}}^4+{2\mathrm{x}}^3+{3\mathrm{x}}^2+2\mathrm{x}+1}=\sqrt{{({\mathrm{x}}^2+\mathrm{x}+1)}^2}$$$$=\mid {\mathrm{x}}^2+\mathrm{x}+1\mid >0\mathrm{for}-\frac{1}{2}⩽\mathrm{x}⩽-\frac{1}{4}$$$$$$

Answered by Dwaipayan Shikari last updated on 12/Jan/21

$${\int }_{-\frac{1}{2}}^{-\frac{1}{4}}\mid x^2+x+1\mid dx$$$$=-(\frac{7}{192}-\frac{3}{32}+\frac{1}{4})=-\frac{37}{192}$$

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