integral of 1/(6+2x)

解答

\int \frac{1}{6 + 2 x} d x

\frac{1}{2} ln ∣ 6 + 2 x ∣ + C

求解步骤

\int \frac{1}{6 + 2 x} d x

使用换元积分法

= \int \frac{1}{2 u} 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} d u

使用积分常见定则:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

u = 6 + 2 x

代回

= \frac{1}{2} ln ∣ 6 + 2 x ∣

解答补常数

= \frac{1}{2} ln ∣ 6 + 2 x ∣ + C

\frac{\partial}{\partial x} (cos (x y) sin (2 y))

d er i v a t i v e x - 2 ln (x + 1)

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

\int_{1}^{4} π (x - 1) d x

d er i v a t i v e 2 (2 - 3 x)^{2}