integral from 0 to 1 of 2/(3+x)

解答

\int_{0}^{1} \frac{2}{3 + x} d x

2 (2 ln (2) - ln (3))

十进制

0.57536 \dots

求解步骤

\int_{0}^{1} \frac{2}{3 + x} d x

提出常数:

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

= 2 \cdot \int_{0}^{1} \frac{1}{3 + x} d x

使用换元积分法

= 2 \cdot \int_{3}^{4} \frac{1}{u} d u

使用积分常见定则:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 [ln ∣ u ∣]_{3}^{4}

带入上下限计算后得:

2 ln (2) - ln (3)

= 2 (2 ln (2) - ln (3))

\int_{8}^{27} (3 x - \frac{1}{3 x}) d x

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

\int_{1}^{2} \frac{4}{2} d x

\int_{0}^{1} x x + 3 d x

\int_{0}^{4} \frac{3}{16} x d x