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受欢迎的微积分 >

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

解答

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

解答

2 - 1

+1

十进制

0.41421 \dots

求解步骤

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

使用换元积分法

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

使用幂法则

= \frac{1}{2} [2 u^{\frac{1}{2}}]_{1}^{2}

使用根式运算法则:

a^{\frac{1}{n}} = n a

= \frac{1}{2} [2 u]_{1}^{2}

带入上下限计算后得:

22 - 2

= \frac{1}{2} (22 - 2)

化简

= 2 - 1

作图

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