integral from 0 to a of 1/(2x+1)

解答

\int_{0}^{a} \frac{1}{2 x + 1} d x

\frac{1}{2} ln ∣ 2 a + 1 ∣

求解步骤

\int_{0}^{a} \frac{1}{2 x + 1} d x

使用换元积分法

= \int_{1}^{2 a + 1} \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_{1}^{2 a + 1} \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 ∣]_{1}^{2 a + 1}

带入上下限计算后得:

ln ∣ 2 a + 1 ∣

= \frac{1}{2} ln ∣ 2 a + 1 ∣