Question Number 212108 by Ismoiljon_008 last updated on 01/Oct/24
$$ \\ $$$$\:\:\:\int_{\:−\mathrm{1}} ^{\:\:\mathrm{1}} \mid\:{x}\:\mid\:\centerdot\:{ln}\left({x}^{\mathrm{2}} \:−\:{x}\:+\:\mathrm{1}\right)\:{dx}\:=\:? \\ $$$$\:\:\:\mathcal{H}{elp}\:{me},\:{please} \\ $$$$ \\ $$
Answered by Frix last updated on 01/Oct/24
![(1) ∫_(−1) ^1 ∣x∣ln (x^2 −x+1) dx= =−∫_(−1) ^0 xln (x^2 −x+1) dx+∫_0 ^1 xln (x^2 −x+1) dx (2) ∫xln (x^2 −x+1) dx =^([t=(((2x−1))/( (√3)))]) =(1/4)∫(3t+(√3))ln ((3(t^2 +1))/4) dt=I+J+K (3) I=(((√3)ln (3/4))/4)∫((√3)t+1)dt=((ln (3/4))/8)t(3t+2(√3))= =((ln (3/4))/8)(2x−1)(2x+1) (4) J=(3/4)∫tln (t^2 +1) dt =^([u=t^2 +1]) (3/8)∫ln u du=((3u)/8)(ln u −1)= =(((x^2 −x+1)ln (x^2 −x+1))/2)+((1+ln (3/4))/2)x(1−x) (5) K=((√3)/4)∫ln (t^2 +1) dt =^([by parts]) =((√3)/4)tln (t^2 +1) −((√3)/2)∫(1−(1/(t^2 +1)))dt= =((√3)/4)(tln (t^2 +1) −2(t−tan^(−1) t))= =(((2x−1)ln (x^2 −x+1))/4)+((√3)/2)tan^(−1) ((2x−1)/( (√3))) −((2+ln (3/4))/2)x (6) ∫xln (x^2 −x+1) dx=I+J+K= =−((x(x+1))/2)+(((2x^2 +1)ln (x^2 −x+1))/4)+((√3)/2)tan^(−1) ((2x−1)/( (√3))) +C (7) Answer is ((π(√3))/(12))+((3ln 3)/4)−1](https://www.tinkutara.com/question/Q212110.png)
$$\left(\mathrm{1}\right) \\ $$$$\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\mid{x}\mid\mathrm{ln}\:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:{dx}= \\ $$$$=−\underset{−\mathrm{1}} {\overset{\mathrm{0}} {\int}}{x}\mathrm{ln}\:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:{dx}+\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}\mathrm{ln}\:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:{dx} \\ $$$$\left(\mathrm{2}\right) \\ $$$$\int{x}\mathrm{ln}\:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:{dx}\:\overset{\left[{t}=\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)}{\:\sqrt{\mathrm{3}}}\right]} {=} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\left(\mathrm{3}{t}+\sqrt{\mathrm{3}}\right)\mathrm{ln}\:\frac{\mathrm{3}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{4}}\:{dt}={I}+{J}+{K} \\ $$$$\left(\mathrm{3}\right) \\ $$$${I}=\frac{\sqrt{\mathrm{3}}\mathrm{ln}\:\frac{\mathrm{3}}{\mathrm{4}}}{\mathrm{4}}\int\left(\sqrt{\mathrm{3}}{t}+\mathrm{1}\right){dt}=\frac{\mathrm{ln}\:\frac{\mathrm{3}}{\mathrm{4}}}{\mathrm{8}}{t}\left(\mathrm{3}{t}+\mathrm{2}\sqrt{\mathrm{3}}\right)= \\ $$$$=\frac{\mathrm{ln}\:\frac{\mathrm{3}}{\mathrm{4}}}{\mathrm{8}}\left(\mathrm{2}{x}−\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{1}\right) \\ $$$$\left(\mathrm{4}\right) \\ $$$${J}=\frac{\mathrm{3}}{\mathrm{4}}\int{t}\mathrm{ln}\:\left({t}^{\mathrm{2}} +\mathrm{1}\right)\:{dt}\:\overset{\left[{u}={t}^{\mathrm{2}} +\mathrm{1}\right]} {=}\:\frac{\mathrm{3}}{\mathrm{8}}\int\mathrm{ln}\:{u}\:{du}=\frac{\mathrm{3}{u}}{\mathrm{8}}\left(\mathrm{ln}\:{u}\:−\mathrm{1}\right)= \\ $$$$=\frac{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\mathrm{ln}\:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{\mathrm{2}}+\frac{\mathrm{1}+\mathrm{ln}\:\frac{\mathrm{3}}{\mathrm{4}}}{\mathrm{2}}{x}\left(\mathrm{1}−{x}\right) \\ $$$$\left(\mathrm{5}\right) \\ $$$${K}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\int\mathrm{ln}\:\left({t}^{\mathrm{2}} +\mathrm{1}\right)\:{dt}\:\overset{\left[\mathrm{by}\:\mathrm{parts}\right]} {=} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{t}\mathrm{ln}\:\left({t}^{\mathrm{2}} +\mathrm{1}\right)\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\int\left(\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}\right){dt}= \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left({t}\mathrm{ln}\:\left({t}^{\mathrm{2}} +\mathrm{1}\right)\:−\mathrm{2}\left({t}−\mathrm{tan}^{−\mathrm{1}} \:{t}\right)\right)= \\ $$$$=\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)\mathrm{ln}\:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{\mathrm{4}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:−\frac{\mathrm{2}+\mathrm{ln}\:\frac{\mathrm{3}}{\mathrm{4}}}{\mathrm{2}}{x} \\ $$$$\left(\mathrm{6}\right) \\ $$$$\int{x}\mathrm{ln}\:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:{dx}={I}+{J}+{K}= \\ $$$$=−\frac{{x}\left({x}+\mathrm{1}\right)}{\mathrm{2}}+\frac{\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{ln}\:\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{\mathrm{4}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+{C} \\ $$$$\left(\mathrm{7}\right) \\ $$$$\mathrm{Answer}\:\mathrm{is} \\ $$$$\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{12}}+\frac{\mathrm{3ln}\:\mathrm{3}}{\mathrm{4}}−\mathrm{1} \\ $$
Commented by Ismoiljon_008 last updated on 01/Oct/24
$$\:\:\:\mathscr{T}\:{hank}\:{you}\:{very}\:{much} \\ $$$$ \\ $$
Commented by Frix last updated on 01/Oct/24
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