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Question Number 182279 by meetbhavsar25 last updated on 06/Dec/22

$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{10}} \left(\mathrm{1}+{x}^{\mathrm{5}} \right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{27}} }{dx}\:=\:? \\$$

Answered by Frix last updated on 07/Dec/22

$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{x}^{\mathrm{10}} \left({x}^{\mathrm{5}} +\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{27}} }{dx}\overset{{t}={x}+\mathrm{1}} {=} \\$$$$=\underset{\mathrm{1}} {\overset{\infty} {\int}}\left(\frac{\mathrm{1}}{{t}^{\mathrm{12}} }−\frac{\mathrm{15}}{{t}^{\mathrm{13}} }+\frac{\mathrm{105}}{{t}^{\mathrm{14}} }−\frac{\mathrm{455}}{{t}^{\mathrm{15}} }+\frac{\mathrm{1365}}{{t}^{\mathrm{16}} }−\frac{\mathrm{3002}}{{t}^{\mathrm{17}} }+\frac{\mathrm{4995}}{{t}^{\mathrm{18}} }−\frac{\mathrm{6390}}{{t}^{\mathrm{19}} }+\frac{\mathrm{6315}}{{t}^{\mathrm{20}} }−\frac{\mathrm{4795}}{{t}^{\mathrm{21}} }+\frac{\mathrm{2751}}{{t}^{\mathrm{22}} }−\frac{\mathrm{1155}}{{t}^{\mathrm{23}} }+\frac{\mathrm{335}}{{t}^{\mathrm{24}} }−\frac{\mathrm{60}}{{t}^{\mathrm{25}} }+\frac{\mathrm{5}}{{t}^{\mathrm{26}} }\right){dt}= \\$$$$=−\left[\frac{\mathrm{1}}{\mathrm{11}{t}^{\mathrm{11}} }−\frac{\mathrm{5}}{\mathrm{4}{t}^{\mathrm{12}} }+\frac{\mathrm{105}}{\mathrm{13}{t}^{\mathrm{13}} }−\frac{\mathrm{65}}{\mathrm{2}{t}^{\mathrm{14}} }+\frac{\mathrm{91}}{{t}^{\mathrm{15}} }−\frac{\mathrm{1501}}{\mathrm{8}{t}^{\mathrm{16}} }+\frac{\mathrm{4995}}{\mathrm{17}{t}^{\mathrm{17}} }−\frac{\mathrm{355}}{{t}^{\mathrm{18}} }+\frac{\mathrm{6315}}{\mathrm{19}{t}^{\mathrm{19}} }−\frac{\mathrm{959}}{\mathrm{4}{t}^{\mathrm{20}} }+\frac{\mathrm{131}}{{t}^{\mathrm{21}} }−\frac{\mathrm{105}}{\mathrm{2}{t}^{\mathrm{22}} }+\frac{\mathrm{335}}{\mathrm{23}{t}^{\mathrm{23}} }−\frac{\mathrm{5}}{\mathrm{2}{t}^{\mathrm{24}} }+\frac{\mathrm{1}}{\mathrm{5}{t}^{\mathrm{25}} }\right]_{\mathrm{1}} ^{\infty} = \\$$$$=\frac{\mathrm{1}}{\mathrm{42}\:\mathrm{493}\:\mathrm{880}} \\$$