Toán tính nguyên hàm sau : A = x^3 × e^(x^2) dx 27/08/2021 By Lyla tính nguyên hàm sau : A = x^3 × e^(x^2) dx
Đáp án: $\begin{array}{l}\int {{x^3}.{e^{{x^2}}}dx} = \int {{x^2}.{e^{{x^2}}}.\frac{1}{2}.2xdx} = \int {\frac{1}{2}.{x^2}.{e^{{x^2}}}d{x^2}} \\Đặt\,{x^2} = u\\ \Rightarrow \int {{x^3}.{e^{{x^2}}}dx} \\ = \int {\frac{1}{2}.u.{e^u}du} \\ = \frac{1}{2}\left( {{e^u}.u – \int {{e^u}du} } \right)\\ = \frac{1}{2}\left( {{e^u}.u – {e^u}} \right) + C\\ = \frac{1}{2}{e^{{x^2}}}\left( {{x^2} – 1} \right) + C\end{array}$ Trả lời
Ta có $A = \int x^3 e^{x^2}dx$ $= \dfrac{1}{2} \int x^2 e^{x^2} d(x^2)$ Đặt $t = x^2$. Khi đó, tích phân trở thành $2A = \int t.e^tdt$ $= \int t.d(e^t)$ $= t.e^t – \int e^t dt$ $= t.e^t – e^t+c$ $= e^t(t-1) +c$ $= e^{x^2}(x^2-1) +c$ Do đó $A = \dfrac{1}{2} e^{x^2} (x^2-1) +c$ Trả lời
Đáp án:
$\begin{array}{l}
\int {{x^3}.{e^{{x^2}}}dx} = \int {{x^2}.{e^{{x^2}}}.\frac{1}{2}.2xdx} = \int {\frac{1}{2}.{x^2}.{e^{{x^2}}}d{x^2}} \\
Đặt\,{x^2} = u\\
\Rightarrow \int {{x^3}.{e^{{x^2}}}dx} \\
= \int {\frac{1}{2}.u.{e^u}du} \\
= \frac{1}{2}\left( {{e^u}.u – \int {{e^u}du} } \right)\\
= \frac{1}{2}\left( {{e^u}.u – {e^u}} \right) + C\\
= \frac{1}{2}{e^{{x^2}}}\left( {{x^2} – 1} \right) + C
\end{array}$
Ta có
$A = \int x^3 e^{x^2}dx$
$= \dfrac{1}{2} \int x^2 e^{x^2} d(x^2)$
Đặt $t = x^2$. Khi đó, tích phân trở thành
$2A = \int t.e^tdt$
$= \int t.d(e^t)$
$= t.e^t – \int e^t dt$
$= t.e^t – e^t+c$
$= e^t(t-1) +c$
$= e^{x^2}(x^2-1) +c$
Do đó
$A = \dfrac{1}{2} e^{x^2} (x^2-1) +c$