integral of (y^2+4y)^2

解答

\int (y^{2} + 4 y)^{2} d y

\frac{y ^{5}}{5} + 2 y^{4} + \frac{16 y ^{3}}{3} + C

求解步骤

\int (y^{2} + 4 y)^{2} d y

乘开

(y^{2} + 4 y)^{2} : y^{4} + 8 y^{3} + 16 y^{2}

= \int y^{4} + 8 y^{3} + 16 y^{2} d y

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

= \int y^{4} d y + \int 8 y^{3} d y + \int 16 y^{2} d y

\int y^{4} d y = \frac{y ^{5}}{5}

\int 8 y^{3} d y = 2 y^{4}

\int 16 y^{2} d y = \frac{16 y ^{3}}{3}

= \frac{y ^{5}}{5} + 2 y^{4} + \frac{16 y ^{3}}{3}

解答补常数

= \frac{y ^{5}}{5} + 2 y^{4} + \frac{16 y ^{3}}{3} + C

d er i v a t i v e 5 x^{3} + 2

\frac{d}{d x} (csc (3 sin (2 + x^{2})))

y^{^{'}} = y x e^{x + 2} - y

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

\frac{\partial}{\partial y} (y + z)