integral from 0 to e of 1/(x+e)

解答

\int_{0}^{e} \frac{1}{x + e} d x

ln (2 e) - 1

十进制

0.69314 \dots

求解步骤

\int_{0}^{e} \frac{1}{x + e} d x

使用换元积分法

= \int_{e}^{2 e} \frac{1}{u} d u

使用积分常见定则:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{e}^{2 e}

带入上下限计算后得:

ln (2 e) - 1

= ln (2 e) - 1

\int \frac{x}{1 + 9 x ^{2}} d x

in v erse l a pl a ce \frac{e ^{- 2 s}}{s - 9}

\int cos^{3} (3 x) sin^{3} (3 x) d x

d er i v a t i v e f (x) = (e^{x} + x^{π}) (x^{2} + x + 1)

y^{^{'}} = 0.1 y (10 - y)