integral of 1/(9x+5)

解答

\int \frac{1}{9 x + 5} d x

\frac{1}{9} ln ∣ 9 x + 5 ∣ + C

求解步骤

\int \frac{1}{9 x + 5} d x

使用换元积分法

= \int \frac{1}{9 u} d u

提出常数:

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

= \frac{1}{9} \cdot \int \frac{1}{u} d u

使用积分常见定则:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{9} ln ∣ u ∣

u = 9 x + 5

代回

= \frac{1}{9} ln ∣ 9 x + 5 ∣

解答补常数

= \frac{1}{9} ln ∣ 9 x + 5 ∣ + C

\frac{\partial}{\partial x} (a x)

d er i v a t i v e 4 ∣ x + 3 ∣ - 5

f (x) = x e^{2 x}

x \to 1 lim (x^{3} - 4 x + 1)

\int \frac{x ^{3}}{9 x ^{2} + 3} d x