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[bqm11111]

Tính

                                            $\int_{0}^{+\infty }\frac{lnxdx}{1-x^{2}}$

Edited by bqm11111, 02-06-2015 - 22:18.

[Mrnhan]

Tính

                                            $\int_{0}^{+\infty }\frac{lnxdx}{1-x^{2}}$

 

$$I=\int_{0}^{\infty}\frac{\ln x}{1-x^2}dx=\int_{0}^{1}\frac{\ln x}{1-x^2}dx+\int_{1}^{\infty}\frac{\ln x}{1-x^2}dx$$

$$=\int_{0}^{1}\frac{\ln x}{1-x^2}dx+\int_{0}^{1}\frac{\ln x}{1-x^2}dx$$

$$=2\int_{0}^{1}\frac{\ln x}{1-x^2}dx$$

$$=\int_{0}^{1}\frac{\ln x}{1-x}+\int_{0}^{1}\frac{\ln x}{1+x}dd$$

 

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[nguyenlyninhkhang]

Lời giải từ : brilliant.org

 

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