bài này em vừa hỏi bạn trí bạn làm thế này ạ
[TEX]I = \int\limits_0^1 {{x^2}\sqrt {1 + {x^2}} } = \int\limits_0^1 {\frac{{{x^2}\left( {{x^2} + 1} \right)}}{{\sqrt {1 + {x^2}} }}} = \int\limits_0^1 {\frac{{{x^2}}}{{\sqrt {{x^2} + 1} }}} + \int\limits_0^1 {\frac{{{x^4}}}{{\sqrt {{x^2} + 1} }}} [/TEX]
[TEX] = \int\limits_0^1 {\frac{{{x^2}}}{{\sqrt {{x^2} + 1} }}} + \left( {\int\limits_0^1 {\frac{{{x^3}x}}{{\sqrt {{x^2} + 1} }}} + \int\limits_0^1 {3{x^2}\sqrt {1 + {x^2}} } } \right) - 3I[/TEX]
[TEX] = \int\limits_0^1 {\frac{{{x^2}}}{{\sqrt {{x^2} + 1} }}} + \left( {{x^3}\sqrt {1 + {x^2}} } \right)\left| \begin{gathered}
^1 \hfill \\
_0 \hfill \\
\end{gathered} \right. - 3I[/TEX]
[TEX] \Rightarrow 4I = \int\limits_0^1 {\frac{{{x^2}}}{{\sqrt {{x^2} + 1} }}} + \sqrt 2 = \int\limits_0^1 {\sqrt {{x^2} + 1} } - \int\limits_0^1 {\frac{1}{{\sqrt {{x^2} + 1} }} + \sqrt 2 } = M - \int\limits_0^1 {\frac{1}{{\sqrt {{x^2} + 1} }}} + \sqrt 2 [/TEX]
[TEX]M = \int\limits_0^1 {\sqrt {{x^2} + 1} } = \int\limits_0^1 {\frac{{{x^2}}}{{\sqrt {{x^2} + 1} }}} + \int\limits_0^1 {\frac{1}{{\sqrt {{x^2} + 1} }}} = \left( {\int\limits_0^1 {\frac{{x.x}}{{\sqrt {{x^2} + 1} }}} + \int\limits_0^1 {\sqrt {{x^2} + 1} } } \right) - M + \int\limits_0^1 {\frac{1}{{\sqrt {{x^2} + 1} }}} [/TEX]
[TEX] = \left( {x\sqrt {1 + {x^2}} } \right)\left| \begin{gathered}
^1 \hfill \\
_0 \hfill \\
\end \right. - M + \int\limits_0^1 {\frac{1}{{\sqrt {{x^2} + 1} }}} [/TEX]
[TEX] \Rightarrow 2M = \sqrt 2 + \int\limits_0^1 {\frac{1}{{\sqrt {{x^2} + 1} }}} [/TEX]
[TEX] \Rightarrow 4I = \frac{1}{2}\left( {\sqrt 2 + \int\limits_0^1 {\frac{1}{{\sqrt {{x^2} + 1} }}} } \right) - \int\limits_0^1 {\frac{1}{{\sqrt {{x^2} + 1} }}} + \sqrt 2 = \frac{{3\sqrt 2 }}{2} - \frac{1}{2}.\int\limits_0^1 {\frac{1}{{\sqrt {{x^2} + 1} }}} = \frac{{3\sqrt 2 }}{2} - \frac{1}{2}.N[/TEX]
các bạn tính nốt N ha
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