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Question Number 27281 by kemhoney78@gmail.com last updated on 04/Jan/18

$$\underset{-1}{\stackrel{1}{\int }}(x^2+\cos x)\log \left(\frac{2+x}{2-x}\right)dx=0$$

Commented by abdo imad last updated on 04/Jan/18

$$letputf\left(x\right)=(x^2+cosx)ln\left(\frac{2+x}{2-x)}\right)wehave$$$$f(-x)=(x^2+cosx)ln\left(\frac{2-x}{2+x}\right)=-(x^2+cosx)ln\left(\frac{2+x}{2-x)}\right)$$$$=-f\left(x\right)fisimpar\Rightarrow {\int }_{-1}^{1}f\left(x\right)dx=0$$

Answered by ajfour last updated on 04/Jan/18

$$I={\int }_{-1}^1(x^2+\cos x)\log \left(\frac{2+x}{2-x}\right)dx$$$$={\int }_{-1}^1(x^2+\cos x)\log \left(\frac{2-x}{2+x}\right)dx$$$$using{\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$$$$2I={\int }_{-1}^1(x^2+\cos x)[\log \left(\frac{2+x}{2-x}\right)+\log \left(\frac{2-x}{2+x}\right)]dx$$$$2I=0\Rightarrow I=0.$$

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