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[TEX]\int_{0}^{1}\sqrt[]{x^2+1}[/TEX]
$\begin{array}{l}
\left\{ \begin{array}{l}
u = \sqrt {{x^2} + 1} \\
dv = dx
\end{array} \right. \to \left\{ \begin{array}{l}
du = \dfrac{x}{{\sqrt {{x^2} + 1} }}dx\\
v = x
\end{array} \right.\\
I = x\sqrt {{x^2} + 1} \left| \begin{array}{l}
1\\
0
\end{array} \right. - \int\limits_0^1 {\dfrac{{{x^2}}}{{\sqrt {{x^2} + 1} }}dx} \\
= \sqrt 2 - \int_0^1 {\sqrt {{x^2} + 1} } dx + \int_0^1 {\dfrac{1}{{\sqrt {{x^2} + 1} }}dx}
\end{array}$
$\begin{array}{l}
\to 2I = \sqrt 2 + \int_0^1 {\dfrac{1}{{\sqrt {{x^2} + 1} }}dx} \\
I' = \int_0^1 {\dfrac{1}{{\sqrt {{x^2} + 1} }}dx} = \int_0^1 {\left( {\dfrac{{x + \sqrt {{x^2} + 1} }}{{\sqrt {{x^2} + 1} }}.\dfrac{1}{{x + \sqrt {{x^2} + 1} }}} \right)dx} \\
I' = \int_0^1 {\dfrac{1}{{x + \sqrt {{x^2} + 1} }}d} \left( {x + \sqrt {{x^2} + 1} } \right) = \ln \left| {x + \sqrt {{x^2} + 1} } \right|\left| \begin{array}{l}
1\\
0
\end{array} \right. = \ln \left( {1 + \sqrt 2 } \right)\\
I = \dfrac{{\sqrt 2 }}{2} + \dfrac{1}{2}\ln \left( {1 + \sqrt 2 } \right)
\end{array}$
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