Đáp án:
\[\frac{5}{3}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
u = {x^2}\\
v' = \frac{1}{{{{\left( {x + 1} \right)}^{\frac{3}{2}}}}} = {\left( {x + 1} \right)^{ - \frac{3}{2}}}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
u' = 2x\\
v = - 2.{\left( {x + 1} \right)^{ - \frac{1}{2}}} = \frac{{ - 2}}{{\sqrt {x + 1} }}
\end{array} \right.\\
I = \int\limits_0^3 {\frac{{{x^2}}}{{{{\left( {x + 1} \right)}^{\frac{3}{2}}}}}dx} = \int\limits_0^3 {{x^2}.\frac{1}{{{{\left( {x + 1} \right)}^{\frac{3}{2}}}}}dx} \\
= \mathop {\left. {{x^2}.\frac{{ - 2}}{{\sqrt {x + 1} }}} \right|}\nolimits_0^3 - \int\limits_0^3 {2x.\frac{{ - 2}}{{\sqrt {x + 1} }}dx} \\
= - 9 + 4\int\limits_0^3 {x.\frac{1}{{\sqrt {x + 1} }}dx} \\
\left\{ \begin{array}{l}
u = x\\
v' = \frac{1}{{\sqrt {x + 1} }} = {\left( {x + 1} \right)^{ - \frac{1}{2}}}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
u' = 1\\
v = 2.{\left( {x + 1} \right)^{\frac{1}{2}}}
\end{array} \right.\\
\Rightarrow I = - 9 + 4.\left( {\mathop {\left. {x.2.{{\left( {x + 1} \right)}^{\frac{1}{2}}}} \right|}\nolimits_0^3 - \int\limits_0^3 {1.2.{{\left( {x + 1} \right)}^{\frac{1}{2}}}dx} } \right)\\
= - 9 + 4.\left( {12 - 2\int\limits_0^3 {{{\left( {x + 1} \right)}^{\frac{1}{2}}}dx} } \right)\\
= - 9 + 4.\left( {12 - 2.\mathop {\left. {\frac{2}{3}.{{\left( {x + 1} \right)}^{\frac{3}{2}}}} \right|}\nolimits_0^3 } \right)\\
= - 9 + 4.\left( {12 - 2.\frac{2}{3}.7} \right) = \frac{5}{3}
\end{array}\)
