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Question-172823




Question Number 172823 by Mikenice last updated on 01/Jul/22
Answered by thfchristopher last updated on 03/Jul/22
∫_0 ^1 cot^(−1) (x^2 −x+1)dx  =[xcot^(−1) (x^2 −x+1)]_0 ^1 +∫_0 ^1 ((x(2x−1))/((x^2 −x+1)^2 +1))dx  =(π/4)+∫_0 ^1 ((2x^2 −x)/(x^4 −2x^3 +3x^2 −2x+2))dx  =(π/4)+∫_0 ^1 ((2x^2 −x)/((x^2 −2x+2)(x^2 +1)))dx  ((2x^2 −x)/((x^2 −2x+2)(x^2 +1)))≡((Ax+B)/(x^2 −2x+2))+((Cx+D)/(x^2 +1))  By solving,  A=1, B=0, C=−1, D=0  ∴ ((2x^2 −x)/((x^2 −2x+2)(x^2 +1)))=(x/(x^2 −2x+2))−(x/(x^2 +1))  ∫_0 ^1 ((2x^2 −x)/((x^2 −2x+2)(x^2 +1)))dx  =∫_0 ^1 (x/(x^2 −2x+2))dx−∫_0 ^1 (x/(x^2 +1))dx  ∫_0 ^1 (x/(x^2 −2x+2))dx  =(1/2)∫_0 ^1 ((2x−2)/(x^2 −2x+2))dx+∫_0 ^1 (dx/(x^2 −2x+2))  =[(1/2)ln (x^2 −2x+2)]_0 ^1 +∫_0 ^1 (dx/((x−1)^2 +1))  =−(1/2)ln 2+[tan^(−1) (x−1)]_0 ^1   =(π/4)−(1/2)ln 2  ∫_0 ^1 (x/(x^2 +1))dx  =(1/2)∫_0 ^1 ((2x)/(x^2 +1))dx  =[(1/2)ln (x^2 +1)]_0 ^1   =(1/2)ln 2  Hence, ∫_0 ^1 cot^(−1) (x^2 −x+1)dx  =(π/4)+(π/4)−(1/2)ln 2−(1/2)ln 2  =(π/2)−ln 2
$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{cot}^{−\mathrm{1}} \left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right){dx} \\ $$$$=\left[{x}\mathrm{cot}^{−\mathrm{1}} \left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}\left(\mathrm{2}{x}−\mathrm{1}\right)}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$=\frac{\pi}{\mathrm{4}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}^{\mathrm{2}} −{x}}{{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}}{dx} \\ $$$$=\frac{\pi}{\mathrm{4}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}^{\mathrm{2}} −{x}}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$$$\frac{\mathrm{2}{x}^{\mathrm{2}} −{x}}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}\equiv\frac{{Ax}+{B}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}}+\frac{{Cx}+{D}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{By}\:\mathrm{solving}, \\ $$$${A}=\mathrm{1},\:{B}=\mathrm{0},\:{C}=−\mathrm{1},\:{D}=\mathrm{0} \\ $$$$\therefore\:\frac{\mathrm{2}{x}^{\mathrm{2}} −{x}}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{x}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}}−\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}^{\mathrm{2}} −{x}}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}−\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}+\left[\mathrm{tan}^{−\mathrm{1}} \left({x}−\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$$\mathrm{Hence},\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{cot}^{−\mathrm{1}} \left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right){dx} \\ $$$$=\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$$=\frac{\pi}{\mathrm{2}}−\mathrm{ln}\:\mathrm{2} \\ $$
Commented by Tawa11 last updated on 04/Jul/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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