$$ \int u^n du = \frac {u^{n+1}}{n+1}, \quad n \neq -1$$
ここで、
$$ u = f(x) $$
$$ du = f'(x) dx $$
とすると、
$$ \int [f(x)]^n f'(x) dx = \frac { [f(x)] ^{n+1}}{n+1}, \quad n \neq -1$$
例
1
$$ \int (3x^2-1)^{\frac {1}{3}} 4x dx $$
\(u = 3x^2-1\)、\(du = 6x dx\)、\(x dx = \frac{1}{6} du\)とすると、
$$ \begin{align} \int (3x^2-1)^{\frac {1}{3}} 4x dx &= \int u^{\frac{1}{3}} \cdot 4 \cdot \frac{1}{6} du \\ &= \frac{2}{3} \int u^{\frac{1}{3}} du \\ &= \frac{2}{3} \cdot \frac{3}{4} u^ {\frac{4}{3}} +c \\ &= \frac{1}{2} u^ {\frac{4}{3}} +c \\ &= \frac{1}{2} (3x^2-1) ^ {\frac{4}{3}} +c \end{align}$$
2
$$ \int \cos 3x dx $$
\(u=3x\)、\(du = 3dx\)、\(dx = \frac{1}{3}du\)とすると、
$$ \begin{align} \int \cos 3x dx &= \int \cos u \cdot \frac{1}{3} du \\ &= \frac{1}{3} \int \cos u du \\ &= \frac{1}{3} \sin u + c \\ &= \frac{1}{3} \sin 3x + c \end{align} $$
3
$$ \int x \sin {(1-x^2)} dx $$
\(u= 1-x^2\)、\(du=-2x dx\)、\(x dx = – \frac{1}{2} du\)とすると、
$$ \begin{align} \int x \sin (1-x^2) dx &= \int \sin u (-\frac{1}{2} du) \\ &= -\frac{1}{2} \int \sin u du \\ &= \frac{1}{2} \cos u + c \\ &= \frac{1}{2} \cos (1-x^2) + c \end{align} $$