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受欢迎的微积分 >

integral from-1 to 1 of 8/(x^2+4)

解答

\int_{- 1}^{1} \frac{8}{x ^{2} + 4} d x

解答

8 arctan (\frac{1}{2})

+1

十进制

3.70918 \dots

求解步骤

\int_{- 1}^{1} \frac{8}{x ^{2} + 4} d x

提出常数:

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

= 8 \cdot \int_{- 1}^{1} \frac{1}{x ^{2} + 4} d x

使用换元积分法

= 8 \cdot \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{1}{2 ( u ^{2} + 1 )} d u

提出常数:

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

= 8 \cdot \frac{1}{2} \cdot \int_{- \frac{1}{2}}^{\frac{1}{2}} \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)

= 8 \cdot \frac{1}{2} [arctan (u)]_{- \frac{1}{2}}^{\frac{1}{2}}

化简

8 \cdot \frac{1}{2} [arctan (u)]_{- \frac{1}{2}}^{\frac{1}{2}} : 4 [arctan (u)]_{- \frac{1}{2}}^{\frac{1}{2}}

= 4 [arctan (u)]_{- \frac{1}{2}}^{\frac{1}{2}}

带入上下限计算后得:

2 arctan (\frac{1}{2})

= 4 \cdot 2 arctan (\frac{1}{2})

化简

= 8 arctan (\frac{1}{2})

作图

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