Giải tích phân bao gồm log ?

J

jet_nguyen

[TEX] I = \int\limits_{1}^{e}x^2log_2xdx[/TEX] .
Ta có: $$(\log_2x)'=\dfrac{1}{x.\ln 2}$$ Nên:

$$I=\int_1^e x^2.\log_2xdx$$$$=\int_1^e \dfrac{\log_2x}{3}dx^3$$$$=\dfrac{x^3\log_2x}{3} \bigg|^e_1 - \int_1^e \dfrac{x^3}{3.x.\ln2}dx$$$$=\dfrac{x^3\log_2x}{3} \bigg|^e_1 - \int_1^e \dfrac{x^2}{3\ln2}dx$$$$=\dfrac{x^3\log_2x}{3} - \dfrac{x^3}{9\ln 2} \bigg|^e_1 $$
 
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