tính tích phân

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nguyenbahiep1

tích phân từng phần thôi

[laTEX]I = \int_{1}^{10} x.log^2xdx \\ \\ u = log^2x \Rightarrow du = 2logx.\frac{1}{xln10}dx \\ \\ dv = x \Rightarrow v = \frac{x^2}{2} \\ \\ I = \frac{x^2log^2x}{2} \big|_1^{10} - \frac{1}{ln10}\int_{1}^{10} xlogxdx \\ \\ I = 100 - \frac{1}{ln10}I_1 \\ \\ I_1 = \frac{x^2logx}{2} \big|_1^{10} - \frac{1}{2ln10}\int_{1}^{10}xdx[/laTEX]
 
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