integral from-1 to 1 of x/(x+2)

解答

\int_{- 1}^{1} \frac{x}{x + 2} d x

2 - 2 ln (3)

十进制

- 0.19722 \dots

求解步骤

\int_{- 1}^{1} \frac{x}{x + 2} d x

使用换元积分法

= \int_{1}^{3} \frac{u - 2}{u} d u

乘开

\frac{u - 2}{u} : 1 - \frac{2}{u}

= \int_{1}^{3} 1 - \frac{2}{u} d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{3} 1 d u - \int_{1}^{3} \frac{2}{u} d u

\int_{1}^{3} 1 d u = 2

\int_{1}^{3} \frac{2}{u} d u = 2 ln (3)

= 2 - 2 ln (3)

\int_{0}^{2} (2 x - 3) (8 x^{2} + 4) d x

\int_{0}^{2} x \cdot sin (\frac{nπ x}{2}) d x

\int_{0}^{2} y e^{x y} d x

\int_{1}^{10} x d x

\int_{1}^{0} (e^{x} + e^{- x}) d x