calculate-2-5-dx-x-1-x-2-

Question Number 38716 by maxmathsup by imad last updated on 28/Jun/18

$${calculate}\:\:\:\int_{\mathrm{2}} ^{\mathrm{5}} \:\:\:\:\:\frac{{dx}}{\left({x}\:+\mathrm{1}−\left[{x}\right]\right)^{\mathrm{2}} } \\ $$

Commented by abdo mathsup 649 cc last updated on 29/Jun/18

$${I}\:=\:\sum_{{k}=\mathrm{2}} ^{\mathrm{4}} \:\int_{{k}} ^{{k}+\mathrm{1}} \:\:\:\frac{{dx}}{\left({x}+\mathrm{1}−{k}\right)^{\mathrm{2}} } \\ $$$$=\sum_{{k}=\mathrm{2}} ^{\mathrm{4}} \:\:\left[−\frac{\mathrm{1}}{{x}+\mathrm{1}−{k}}\right]_{{k}} ^{{k}+\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{2}} ^{\mathrm{4}} \:\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right\}=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{2}} ^{\mathrm{4}} \left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}−\mathrm{2}+\mathrm{1}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jun/18

$$\int_{\mathrm{2}} ^{\mathrm{3}} \frac{{dx}}{\left({x}+\mathrm{1}−\mathrm{2}\right)^{\mathrm{2}} }+\int_{\mathrm{3}} ^{\mathrm{4}} \frac{{dx}}{\left({x}+\mathrm{1}−\mathrm{3}\right)^{\mathrm{2}} }+\int_{\mathrm{4}} ^{\mathrm{5}} \frac{{dx}}{\left({x}+\mathrm{1}−\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$=\mid\frac{\left({x}−\mathrm{1}\right)^{−\mathrm{1}} }{−\mathrm{1}}\mid_{\mathrm{2}} ^{\mathrm{3}} +\mid\frac{\left({x}−\mathrm{2}\right)^{−\mathrm{1}} }{−\mathrm{1}}\mid_{\mathrm{3}} ^{\mathrm{4}} +\mid\frac{\left({x}−\mathrm{3}\right)^{−\mathrm{1}} }{−\mathrm{1}}\mid_{\mathrm{4}} ^{\mathrm{5}} \\ $$$$=−\mathrm{1}\left\{\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{1}}\right)\right\} \\ $$$$=−\mathrm{1}\left(\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{3}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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