Tích phân có log ?

N

newstarinsky

[TEX] I = \int\limits_{1}^{e}x^2log_2xdx[/TEX] .

đặt
[TEX]\left{\begin{u=log_2x}\\{dv=x^2dx} [/TEX]


[TEX]\Rightarrow\left{\begin{du=\frac{1}{x.ln2}}\\{v= x^3/3}[/TEX]

nên [TEX]I=(log_2x.\frac{1}{3}x^3)\left \uparrow \frac{e}{1} -\frac{1}{3ln2}.\int_{1}^{e}x^2dx[/TEX]

Dạng quen rồi
 
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