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Question Number 198136 by sonukgindia last updated on 11/Oct/23

Answered by Frix last updated on 11/Oct/23

$$\frac{3x^3+4x^2+4x+16}{x^5+2x^4+16x+32}=\frac{(x+4)(x^2+4)}{(x+2)(x^4+16)}=$$$$=\frac{1}{2(x+2)}-\frac{(2-\sqrt{2})x+2(2+\sqrt{2})}{8(x^2-2\sqrt{2}x+4)}-\frac{(2+\sqrt{2})x-2(2-\sqrt{2})}{8(x^2+2\sqrt{2}x+4)}$$$$\mathrm{Now}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}.\mathrm{I}\mathrm{get}$$$$I=\frac{3\sqrt{2}}{8}\pi$$

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