integral from 3 to 9 of 1/(9+x^2)

解答

\int_{3}^{9} \frac{1}{9 + x ^{2}} d x

\frac{1}{3} (arctan (3) - \frac{π}{4})

十进制

0.15454 \dots

求解步骤

\int_{3}^{9} \frac{1}{9 + x ^{2}} d x

使用换元积分法

= \int_{1}^{3} \frac{1}{3 ( 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}{3} \cdot \int_{1}^{3} \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}{3} [arctan (u)]_{1}^{3}

带入上下限计算后得:

arctan (3) - \frac{π}{4}

= \frac{1}{3} (arctan (3) - \frac{π}{4})

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