integral from 0 to 1 of x/(x^2+9)

解答

\int_{0}^{1} \frac{x}{x ^{2} + 9} d x

\frac{1}{2} ln (10) - ln (3)

十进制

0.05268 \dots

求解步骤

\int_{0}^{1} \frac{x}{x ^{2} + 9} d x

使用换元积分法

= \int_{9}^{10} \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_{9}^{10} \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 ∣]_{9}^{10}

带入上下限计算后得:

ln (10) - 2 ln (3)

= \frac{1}{2} (ln (10) - 2 ln (3))

化简

= \frac{1}{2} ln (10) - ln (3)

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