integral of (x^2-1)^2(3x^2)

解答

\int (x^{2} - 1)^{2} (3 x^{2}) d x

\frac{3 x ^{7}}{7} - \frac{6 x ^{5}}{5} + x^{3} + C

求解步骤

\int (x^{2} - 1)^{2} \cdot 3 x^{2} d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int (x^{2} - 1)^{2} x^{2} d x

乘开

(x^{2} - 1)^{2} x^{2} : x^{6} - 2 x^{4} + x^{2}

= 3 \cdot \int x^{6} - 2 x^{4} + x^{2} d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int x^{6} d x - \int 2 x^{4} d x + \int x^{2} d x)

\int x^{6} d x = \frac{x ^{7}}{7}

\int 2 x^{4} d x = \frac{2 x ^{5}}{5}

\int x^{2} d x = \frac{x ^{3}}{3}

= 3 (\frac{x ^{7}}{7} - \frac{2 x ^{5}}{5} + \frac{x ^{3}}{3})

化简

3 (\frac{x ^{7}}{7} - \frac{2 x ^{5}}{5} + \frac{x ^{3}}{3}) : \frac{3 x ^{7}}{7} - \frac{6 x ^{5}}{5} + x^{3}

= \frac{3 x ^{7}}{7} - \frac{6 x ^{5}}{5} + x^{3}

解答补常数

= \frac{3 x ^{7}}{7} - \frac{6 x ^{5}}{5} + x^{3} + C

\frac{d x}{d y} = e^{3 x + 2 y}

x \to 3 lim (\frac{0}{x - 3})

y^{^{''}} - 16 y = \frac{16 x}{e ^{4 x}}

y^{^{''}} + 2 y + y = 2 e^{- t}

\frac{d}{d θ} (\frac{1}{4} sin^{2} (2 θ))