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

解答

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

\frac{π}{4 m}

求解步骤

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

x^{2} + m = m (\frac{x ^{2}}{m} + 1)

= \int_{0}^{m} \frac{1}{m ( \frac{x ^{2}}{m} + 1 )} d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

使用换元积分法

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

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

使用积分常见定则:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{m} m [arctan (u)]_{0}^{\frac{1}{m} m}

化简

\frac{1}{m} m [arctan (u)]_{0}^{\frac{1}{m} m} : \frac{1}{m} [arctan (u)]_{0}^{1}

= \frac{1}{m} [arctan (u)]_{0}^{1}

带入上下限计算后得:

\frac{π}{4}

= \frac{1}{m} \cdot \frac{π}{4}

化简

= \frac{π}{4 m}

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