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integral of 9/(x^2sqrt(x^2+36))

解答

\int \frac{9}{x ^{2} x ^{2} + 36} d x

解答

- \frac{36 + x ^{2}}{4 x} + C

求解步骤

\int \frac{9}{x ^{2} x ^{2} + 36} d x

提出常数:

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

= 9 \cdot \int \frac{1}{x ^{2} x ^{2} + 36} d x

使用三角换元法

= 9 \cdot \int \frac{sec ( u )}{36 tan ^{2} ( u )} d u

提出常数:

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

= 9 \cdot \frac{1}{36} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

使用三角恒等式改写

= 9 \cdot \frac{1}{36} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

使用换元积分法

= 9 \cdot \frac{1}{36} \cdot \int \frac{1}{v ^{2}} d v

使用幂法则

= 9 \cdot \frac{1}{36} (- \frac{1}{v})

代回

= 9 \cdot \frac{1}{36} (- \frac{1}{sin ( arctan ( \frac{1}{6} x ) )})

化简

9 \cdot \frac{1}{36} (- \frac{1}{sin ( arctan ( \frac{1}{6} x ) )}) : - \frac{36 + x ^{2}}{4 x}

= - \frac{36 + x ^{2}}{4 x}

解答补常数

= - \frac{36 + x ^{2}}{4 x} + C

作图

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