integral of 1/(2y-1)

解答

\int \frac{1}{2 y - 1} d y

\frac{1}{2} ln ∣ 2 y - 1 ∣ + C

求解步骤

\int \frac{1}{2 y - 1} d y

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 = 2 y - 1

代回

= \frac{1}{2} ln ∣ 2 y - 1 ∣

解答补常数

= \frac{1}{2} ln ∣ 2 y - 1 ∣ + C

t an g e n t y = (x^{3} + 2) (3 x^{2} - 4 x + 2), (1, 3)

t an g e n t x^{2} (2.4)

\int 7 x + 5 d x

\int \frac{( 2 + x )}{1 - x ^{2}} d x

x \to 1 lim (x - 1 x)