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受欢迎的微积分 >

integral from 0 to 2 of 1/((x^2+4)^2)

解答

\int_{0}^{2} \frac{1}{( x ^{2} + 4 ) ^{2}} d x

解答

\frac{π + 2}{64}

+1

十进制

0.08033 \dots

求解步骤

\int_{0}^{2} \frac{1}{( x ^{2} + 4 ) ^{2}} d x

使用换元积分法

= \int_{0}^{1} \frac{1}{8 ( u ^{2} + 1 ) ^{2}} d u

提出常数:

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

= \frac{1}{8} \cdot \int_{0}^{1} \frac{1}{( u ^{2} + 1 ) ^{2}} d u

使用三角换元法

= \frac{1}{8} \cdot \int_{0}^{\frac{π}{4}} \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

使用三角恒等式改写

= \frac{1}{8} \cdot \int_{0}^{\frac{π}{4}} \frac{1 + cos ( 2 v )}{2} d v

提出常数:

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

= \frac{1}{8} \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} 1 + cos (2 v) d v

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} \cdot \frac{1}{2} (\int_{0}^{\frac{π}{4}} 1 d v + \int_{0}^{\frac{π}{4}} cos (2 v) d v)

\int_{0}^{\frac{π}{4}} 1 d v = \frac{π}{4}

\int_{0}^{\frac{π}{4}} cos (2 v) d v = \frac{1}{2}

= \frac{1}{8} \cdot \frac{1}{2} (\frac{π}{4} + \frac{1}{2})

化简

\frac{1}{8} \cdot \frac{1}{2} (\frac{π}{4} + \frac{1}{2}) : \frac{π + 2}{64}

= \frac{π + 2}{64}

作图

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