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Question Number 212108 by Ismoiljon_008 last updated on 01/Oct/24

$$$$$${\int }_{-1}^1\mid x\mid \cdot ln(x^2-x+1)dx=?$$$$\mathcal{H}elpme,please$$$$$$

Answered by Frix last updated on 01/Oct/24

$$\left(1\right)$$$$\underset{-1}{\stackrel{1}{\int }}\mid x\mid \ln (x^2-x+1)dx=$$$$=-\underset{-1}{\stackrel{0}{\int }}x\ln (x^2-x+1)dx+\underset{0}{\stackrel{1}{\int }}x\ln (x^2-x+1)dx$$$$\left(2\right)$$$$\int x\ln (x^2-x+1)dx\stackrel{[t=\frac{(2x-1)}{\sqrt{3}}]}{=}$$$$=\frac{1}{4}\int (3t+\sqrt{3})\ln \frac{3(t^2+1)}{4}dt=I+J+K$$$$\left(3\right)$$$$I=\frac{\sqrt{3}\ln \frac{3}{4}}{4}\int (\sqrt{3}t+1)dt=\frac{\ln \frac{3}{4}}{8}t(3t+2\sqrt{3})=$$$$=\frac{\ln \frac{3}{4}}{8}(2x-1)(2x+1)$$$$\left(4\right)$$$$J=\frac{3}{4}\int t\ln (t^2+1)dt\stackrel{[u=t^2+1]}{=}\frac{3}{8}\int \ln udu=\frac{3u}{8}(\ln u-1)=$$$$=\frac{(x^2-x+1)\ln (x^2-x+1)}{2}+\frac{1+\ln \frac{3}{4}}{2}x(1-x)$$$$\left(5\right)$$$$K=\frac{\sqrt{3}}{4}\int \ln (t^2+1)dt\stackrel{\left[\mathrm{by}\mathrm{parts}\right]}{=}$$$$=\frac{\sqrt{3}}{4}t\ln (t^2+1)-\frac{\sqrt{3}}{2}\int (1-\frac{1}{t^2+1})dt=$$$$=\frac{\sqrt{3}}{4}(t\ln (t^2+1)-2(t-{\tan }^{-1}t))=$$$$=\frac{(2x-1)\ln (x^2-x+1)}{4}+\frac{\sqrt{3}}{2}{\tan }^{-1}\frac{2x-1}{\sqrt{3}}-\frac{2+\ln \frac{3}{4}}{2}x$$$$\left(6\right)$$$$\int x\ln (x^2-x+1)dx=I+J+K=$$$$=-\frac{x(x+1)}{2}+\frac{(2x^2+1)\ln (x^2-x+1)}{4}+\frac{\sqrt{3}}{2}{\tan }^{-1}\frac{2x-1}{\sqrt{3}}+C$$$$\left(7\right)$$$$\mathrm{Answer}\mathrm{is}$$$$\frac{\pi \sqrt{3}}{12}+\frac{3\ln 3}{4}-1$$

Commented by Ismoiljon_008 last updated on 01/Oct/24

$$\mathcal{T}hankyouverymuch$$$$$$

Commented by Frix last updated on 01/Oct/24

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