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Question Number 57418 by Abdo msup. last updated on 03/Apr/19

calculate ∫_(−1) ^4 ((∣x−1∣+∣x−2∣)/(∣x^2 −9∣ +x^2 +16))dx

$${calculate}\:\int_{−\mathrm{1}} ^{\mathrm{4}} \:\frac{\mid{x}−\mathrm{1}\mid+\mid{x}−\mathrm{2}\mid}{\mid{x}^{\mathrm{2}} −\mathrm{9}\mid\:+{x}^{\mathrm{2}} \:+\mathrm{16}}{dx} \\ $$

Commented by kaivan.ahmadi last updated on 03/Apr/19

∫_(−1) ^1 ((1−x+2−x)/(9−x^2 +x^2 +16))dx+∫_1 ^2 ((x−1+2−x)/(9−x^2 +x^2 +16))dx+ ∫_2 ^3 ((x+1+x+2)/(9−x^2 +x^2 +16))dx+∫_3 ^4 ((x−1+x+2)/(x^2 −9+x^2 +16))dx= ∫_(−1) ^1 ((−2x+3)/(25))dx+∫_1 ^2 (1/(25))dx+ ∫_2 ^3 ((2x+3)/(25))dx+∫_3 ^4 ((2x+1)/(2x^2 +7))dx now you can calculate each of this integral easily

$$\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}+\mathrm{2}−{x}}{\mathrm{9}−{x}^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{16}}{dx}+\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{x}−\mathrm{1}+\mathrm{2}−{x}}{\mathrm{9}−{x}^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{16}}{dx}+ \\ $$$$\int_{\mathrm{2}} ^{\mathrm{3}} \frac{{x}+\mathrm{1}+{x}+\mathrm{2}}{\mathrm{9}−{x}^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{16}}{dx}+\int_{\mathrm{3}} ^{\mathrm{4}} \frac{{x}−\mathrm{1}+{x}+\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{9}+{x}^{\mathrm{2}} +\mathrm{16}}{dx}= \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{−\mathrm{2}{x}+\mathrm{3}}{\mathrm{25}}{dx}+\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{25}}{dx}+ \\ $$$$\int_{\mathrm{2}} ^{\mathrm{3}} \frac{\mathrm{2}{x}+\mathrm{3}}{\mathrm{25}}{dx}+\int_{\mathrm{3}} ^{\mathrm{4}} \frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{7}}{dx} \\ $$$${now}\:{you}\:{can}\:{calculate}\:{each}\:{of} \\ $$$${this}\:{integral}\:{easily} \\ $$

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