integral from 0 to 9/7 of 1/(49x^2+81)

解答

\int_{0}^{\frac{9}{7}} \frac{1}{49 x ^{2} + 81} d x

\frac{π}{252}

十进制

0.01246 \dots

求解步骤

\int_{0}^{\frac{9}{7}} \frac{1}{49 x ^{2} + 81} d x

使用换元积分法

= \int_{0}^{1} \frac{1}{63 ( 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}{63} \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 = arctan (u)

= \frac{1}{63} [arctan (u)]_{0}^{1}

带入上下限计算后得:

\frac{π}{4}

= \frac{1}{63} \cdot \frac{π}{4}

化简

= \frac{π}{252}

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