integral of 2/(x+1)

解答

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

2 ln ∣ x + 1 ∣ + C

求解步骤

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

提出常数:

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

= 2 \cdot \int \frac{1}{x + 1} d x

使用换元积分法

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

使用积分常见定则:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 ln ∣ u ∣

u = x + 1

代回

= 2 ln ∣ x + 1 ∣

解答补常数

= 2 ln ∣ x + 1 ∣ + C

y^{^{'}} = \frac{x y ^{2} - 0.5 sin ( 2 x )}{y ( 1 - x ^{2} )}, y (0) = 2

y^{^{'}} = (t + y)^{2} - 1, y (2) = 6

\frac{d}{d x} ((\frac{x + 2}{x - 2})^{5})

\frac{d}{d x} (2 x^{2} + y^{2})

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