Tìm nguyên hàm $\int\ {\frac{x^3}{\sqrt{2- x^2}}} \, dx$ 25/07/2021 Bởi Margaret Tìm nguyên hàm $\int\ {\frac{x^3}{\sqrt{2- x^2}}} \, dx$
Đáp án: $\begin{array}{l}I = \int {\frac{{{x^3}}}{{\sqrt {2 – {x^2}} }}dx} \\ = \int {\frac{{{x^2}.2x}}{{2\sqrt {2 – {x^2}} }}dx} \\ = \int {\frac{{{x^2}}}{{2\sqrt {2 – {x^2}} }}d\left( {{x^2}} \right)} \\Đặt\,\sqrt {2 – {x^2}} = u\\ \Rightarrow 2 – {x^2} = {u^2}\\ \Rightarrow \left\{ \begin{array}{l}d\left( {{x^2}} \right) = – 2udu\\{x^2} = 2 – {u^2}\end{array} \right.\\ \Rightarrow I = \int {\frac{{2 – {u^2}}}{{2u}}.\left( { – 2u} \right)du} \\ = \int {\left( {{u^2} – 2} \right)du} \\ = \frac{1}{3}{u^3} – 2u + C\\ = \frac{1}{3}.\left( {2 – {x^2}} \right)\sqrt {2 – {x^2}} – 2\sqrt {2 – {x^2}} + C\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
I = \int {\frac{{{x^3}}}{{\sqrt {2 – {x^2}} }}dx} \\
= \int {\frac{{{x^2}.2x}}{{2\sqrt {2 – {x^2}} }}dx} \\
= \int {\frac{{{x^2}}}{{2\sqrt {2 – {x^2}} }}d\left( {{x^2}} \right)} \\
Đặt\,\sqrt {2 – {x^2}} = u\\
\Rightarrow 2 – {x^2} = {u^2}\\
\Rightarrow \left\{ \begin{array}{l}
d\left( {{x^2}} \right) = – 2udu\\
{x^2} = 2 – {u^2}
\end{array} \right.\\
\Rightarrow I = \int {\frac{{2 – {u^2}}}{{2u}}.\left( { – 2u} \right)du} \\
= \int {\left( {{u^2} – 2} \right)du} \\
= \frac{1}{3}{u^3} – 2u + C\\
= \frac{1}{3}.\left( {2 – {x^2}} \right)\sqrt {2 – {x^2}} – 2\sqrt {2 – {x^2}} + C
\end{array}$