integral of 1/(7+9x^2)

解答

\int \frac{1}{7 + 9 x ^{2}} d x

\frac{1}{3 7} arctan (\frac{3}{7} x) + C

求解步骤

\int \frac{1}{7 + 9 x ^{2}} d x

使用换元积分法

= \int \frac{1}{3 7 ( 1 + u ^{2} )} d u

提出常数:

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

= \frac{1}{3 7} \cdot \int \frac{1}{1 + u ^{2}} d u

使用积分常见定则:

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

= \frac{1}{3 7} arctan (u)

u = \frac{3}{7} x

代回

= \frac{1}{3 7} arctan (\frac{3}{7} x)

解答补常数

= \frac{1}{3 7} arctan (\frac{3}{7} x) + C

(2 x - 1) d x + (3 x + 7) d y = 0

\frac{d}{d x} (\frac{1}{I n ( x + 2 )})

\int [5 x + \frac{2}{3 x ^{5}}] d x

d er i v a t i v e 2 a x + \frac{b}{x}

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