integral of 1/(y^2+2)

解答

\int \frac{1}{y ^{2} + 2} d y

\frac{1}{2} arctan (\frac{y}{2}) + C

求解步骤

\int \frac{1}{y ^{2} + 2} d y

使用换元积分法

= \int \frac{1}{2 ( u ^{2} + 1 )} d u

提出常数:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 1} d u

使用积分常见定则:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{2} arctan (u)

u = \frac{y}{2}

代回

= \frac{1}{2} arctan (\frac{y}{2})

解答补常数

= \frac{1}{2} arctan (\frac{y}{2}) + C

\frac{\partial}{\partial x} (12 - x^{2} - y^{2})

y^{^{'}} = k y (100 - y)

\frac{d}{d x} (11 sin (x))

\frac{\partial}{\partial x} (2 y + z^{3} - x z)

d er i v a t i v e y = \frac{2 x}{x + 1}