integral from 0 to 5 of x/(x^2+1)

解答

\int_{0}^{5} \frac{x}{x ^{2} + 1} d x

\frac{1}{2} ln (26)

十进制

1.62904 \dots

求解步骤

\int_{0}^{5} \frac{x}{x ^{2} + 1} d x

使用换元积分法

= \int_{1}^{26} \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}^{26} \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}^{26}

带入上下限计算后得:

ln (26)

= \frac{1}{2} ln (26)

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