Tích phân hàm hữu tỉ :

G

giahung341_14

[tex]I = \int\limits_0^1 {\frac{{{x^2}}}{{{{(3x + 1)}^3}}}dx} [/tex]
[tex]t = 3x + 1 \Rightarrow dt = 3dx[/tex]
[tex] \Rightarrow x = \frac{{t - 1}}{3}[/tex]
[tex]x = 0 \Rightarrow t = 1[/tex]
[tex]x = 1 \Rightarrow t = 4[/tex]
[tex]I = {\int\limits_1^4 {\left( {\frac{{t - 1}}{3}} \right)} ^2}.\frac{1}{{{t^3}}}dt = \frac{1}{9}\int\limits_1^4 {\left( {\frac{{{t^2} - 2t + 1}}{{{t^3}}}} \right)} dt = \frac{1}{9}\int\limits_1^4 {\left( {\frac{1}{t} - \frac{2}{{{t^2}}} + \frac{1}{{{t^3}}}} \right)} dt[/tex]
Tới đây bạn làm tiếp nha
 
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