Tìm nguyên hàm của f(x) = $\sqrt{3x+2}$

$@Vân$

$\int {f\left( x \right)dx} = \frac{1}{{3.\left( {1 + \frac{1}{2}} \right)}}.{\left( {\sqrt {3x + 2} } \right)^{1 + \frac{1}{2}}} + c$

$= \frac{2}{9}\left( {3x + 2} \right)\sqrt {3x + 2} + c$

Đáp án:

\(\displaystyle\int\sqrt{3x + 2}dx =\dfrac29(3x+2)\sqrt{3x+2} + C\)

Giải thích các bước giải:

\(\begin{array}{l}
\quad \displaystyle\int f(x)dx = \displaystyle\int\sqrt{3x + 2}dx\\
\to \displaystyle\int f(x)dx = \dfrac13\displaystyle\int\sqrt{3x + 2}d(3x + 2)\\
\to \displaystyle\int f(x)dx = \dfrac13\cdot \dfrac23\sqrt{(3x+2)^3} + C\\
\to\displaystyle\int f(x)dx = \dfrac29(3x+2)\sqrt{3x+2} + C
\end{array}\)