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受欢迎的微积分 >

integral of 8/(x^2sqrt(x^2+4))

解答

\int \frac{8}{x ^{2} x ^{2} + 4} d x

解答

- \frac{2 x ^{2} + 4}{x} + C

求解步骤

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

使用三角换元法

= 8 \cdot \int \frac{sec ( u )}{4 tan ^{2} ( u )} d u

提出常数:

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

= 8 \cdot \frac{1}{4} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

使用三角恒等式改写

= 8 \cdot \frac{1}{4} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

使用换元积分法

= 8 \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{2}} d v

使用幂法则

= 8 \cdot \frac{1}{4} (- \frac{1}{v})

代回

= 8 \cdot \frac{1}{4} (- \frac{1}{sin ( arctan ( \frac{1}{2} x ) )})

化简

8 \cdot \frac{1}{4} (- \frac{1}{sin ( arctan ( \frac{1}{2} x ) )}) : - \frac{2 x ^{2} + 4}{x}

= - \frac{2 x ^{2} + 4}{x}

解答补常数

= - \frac{2 x ^{2} + 4}{x} + C

作图

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