Question Number 198136 by sonukgindia last updated on 11/Oct/23
Answered by Frix last updated on 11/Oct/23
$$\frac{\mathrm{3}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{16}}{{x}^{\mathrm{5}} +\mathrm{2}{x}^{\mathrm{4}} +\mathrm{16}{x}+\mathrm{32}}=\frac{\left({x}+\mathrm{4}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{\left({x}+\mathrm{2}\right)\left({x}^{\mathrm{4}} +\mathrm{16}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{2}\right)}−\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){x}+\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{8}\left({x}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}{x}+\mathrm{4}\right)}−\frac{\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){x}−\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}{\mathrm{8}\left({x}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}{x}+\mathrm{4}\right)} \\ $$$$\mathrm{Now}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}.\:\mathrm{I}\:\mathrm{get} \\ $$$${I}=\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{8}}\pi \\ $$