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

解答

\int_{0}^{1} \frac{1}{x ( x + 1 )} d x

2 ln (1 + 2)

十进制

1.76274 \dots

求解步骤

\int_{0}^{1} \frac{1}{x ( x + 1 )} d x

使用指数法则:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

x (x + 1) = x (x + 1)

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

使用换元积分法

= \int_{0}^{1} \frac{2}{u ^{2} + 1} d u

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}{u ^{2} + 1} d u

使用积分常见定则:

\int \frac{1}{u ^{2} + 1} d u = ln u^{2} + 1 + u

= 2 [ln u^{2} + 1 + u]_{0}^{1}

带入上下限计算后得:

ln (1 + 2)

= 2 ln (1 + 2)

\int_{- 2}^{- 2} 1

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