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

解答

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

\frac{π}{36}

十进制

0.08726 \dots

求解步骤

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

使用换元积分法

= \int_{0}^{1} \frac{1}{9 ( 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}{9} \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}{9} [arctan (u)]_{0}^{1}

带入上下限计算后得:

\frac{π}{4}

= \frac{1}{9} \cdot \frac{π}{4}

化简

= \frac{π}{36}

\int_{0}^{l n (4)} 6 e^{2 x} d x

\int_{0}^{4 π} sin (2 x) d x

\int_{- 3}^{1} - x^{2} - 2 x + 3 d x

\int_{0}^{9} ∣ 7 - x ∣ d x

\int_{0}^{2} x (2 - x) d x