tìm nguyên hàm $\int\limits{x^5 \sqrt[]{x^2+9} } \, dx$ 24/07/2021 Bởi Arya tìm nguyên hàm $\int\limits{x^5 \sqrt[]{x^2+9} } \, dx$
Đáp án: \(\dfrac{1}{35}\sqrt{\left(x^2 + 9\right)^3}\left(5x^4 – 36x^2 + 216\right) + C\) Giải thích các bước giải: \(\begin{array}{l}\quad I = \displaystyle\int x^5\sqrt{x^2 + 9}dx\\\text{Đặt}\ u = x^2 + 9\\\Rightarrow du = 2xdx\\\text{Ta được:}\\\quad I = \dfrac12\displaystyle\int (u – 9)^2\sqrt udu\\\Leftrightarrow I = \dfrac12\displaystyle\int\left(u^{\tfrac52} – 18u^{\tfrac32} + 81u^{\tfrac12}\right)du\\\Leftrightarrow I = \dfrac12\left(\dfrac27u^{\tfrac72} – \dfrac{36}{5}u^{\tfrac52} + 54u^{\tfrac32} \right) + C\\\Leftrightarrow I = u^{\tfrac32}\left(\dfrac17u^2 – \dfrac{18}{5}u + 27\right) + C\\\Leftrightarrow I = \left(x^2 + 9\right)^{\tfrac32}\left[\dfrac17\left(x^2 + 9\right)^2 – \dfrac{18}{5}\left(x^2 + 9\right) + 27\right] + C\\\Leftrightarrow I = \dfrac{1}{35}\sqrt{\left(x^2 + 9\right)^3}\left(5x^4 – 36x^2 + 216\right) + C\end{array}\) Bình luận
Đáp án:
\(\dfrac{1}{35}\sqrt{\left(x^2 + 9\right)^3}\left(5x^4 – 36x^2 + 216\right) + C\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad I = \displaystyle\int x^5\sqrt{x^2 + 9}dx\\
\text{Đặt}\ u = x^2 + 9\\
\Rightarrow du = 2xdx\\
\text{Ta được:}\\
\quad I = \dfrac12\displaystyle\int (u – 9)^2\sqrt udu\\
\Leftrightarrow I = \dfrac12\displaystyle\int\left(u^{\tfrac52} – 18u^{\tfrac32} + 81u^{\tfrac12}\right)du\\
\Leftrightarrow I = \dfrac12\left(\dfrac27u^{\tfrac72} – \dfrac{36}{5}u^{\tfrac52} + 54u^{\tfrac32} \right) + C\\
\Leftrightarrow I = u^{\tfrac32}\left(\dfrac17u^2 – \dfrac{18}{5}u + 27\right) + C\\
\Leftrightarrow I = \left(x^2 + 9\right)^{\tfrac32}\left[\dfrac17\left(x^2 + 9\right)^2 – \dfrac{18}{5}\left(x^2 + 9\right) + 27\right] + C\\
\Leftrightarrow I = \dfrac{1}{35}\sqrt{\left(x^2 + 9\right)^3}\left(5x^4 – 36x^2 + 216\right) + C
\end{array}\)