Question Number 198948 by ArifinTanjung last updated on 26/Oct/23 | ||
$$\int_{\mathrm{1}} ^{\mathrm{3}} \frac{\mathrm{x}−\mathrm{2}}{\mathrm{x}^{\mathrm{2}} −\mathrm{4x}}\:\mathrm{dx}=\:.... \\ $$ | ||
Answered by ajfour last updated on 26/Oct/23 | ||
$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{1}} ^{\:\mathrm{1}} \frac{\mathrm{2}{tdt}}{{t}^{\mathrm{2}} −\mathrm{4}}=\mathrm{0}\:\:\:\:\:\left({odd}\:{function}\right) \\ $$$$\int_{−{a}} ^{\:{a}} {f}\left({x}\right){dx}=\mathrm{0}\:\:{if}\:\:{f}\left(−{x}\right)=−{f}\left({x}\right) \\ $$ | ||
Answered by mr W last updated on 26/Oct/23 | ||
$$\int_{\mathrm{1}} ^{\mathrm{3}} \frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{4}{x}}{dx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{4}{x}}{dx}+\int_{\mathrm{2}} ^{\mathrm{3}} \frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{4}{x}}{dx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{4}{x}}{dx}+\int_{\mathrm{2}} ^{\mathrm{1}} \frac{\mathrm{4}−{t}−\mathrm{2}}{\left(\mathrm{4}−{t}\right)\left(\mathrm{4}−{t}−\mathrm{4}\right)}{d}\left(\mathrm{4}−{t}\right) \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{4}{x}}{dx}+\int_{\mathrm{2}} ^{\mathrm{1}} \frac{{t}−\mathrm{2}}{{t}^{\mathrm{2}} −\mathrm{4}{t}}{dt} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{4}{x}}{dx}−\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{4}{x}}{dx} \\ $$$$=\mathrm{0} \\ $$ | ||