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受欢迎的微积分 >

integral of-1/(2xsqrt(x^2+9))

解答

\int - \frac{1}{2 x x ^{2} + 9} d x

解答

- \frac{1}{6} ln tan (\frac{1}{2} arctan (\frac{1}{3} x)) + C

求解步骤

\int - \frac{1}{2 x x ^{2} + 9} d x

提出常数:

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

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

使用三角换元法

= - \frac{1}{2} \cdot \int \frac{sec ( u )}{3 tan ( u )} d u

提出常数:

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

= - \frac{1}{2} \cdot \frac{1}{3} \cdot \int \frac{sec ( u )}{tan ( u )} d u

使用换元积分法

= - \frac{1}{2} \cdot \frac{1}{3} \cdot \int \frac{1}{v} d v

使用积分常见定则:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - \frac{1}{2} \cdot \frac{1}{3} ln ∣ v ∣

代回

= - \frac{1}{2} \cdot \frac{1}{3} ln tan (\frac{arctan ( \frac{1}{3} x )}{2})

化简

- \frac{1}{2} \cdot \frac{1}{3} ln tan (\frac{arctan ( \frac{1}{3} x )}{2}) : - \frac{1}{6} ln tan (\frac{1}{2} arctan (\frac{1}{3} x))

= - \frac{1}{6} ln tan (\frac{1}{2} arctan (\frac{1}{3} x))

解答补常数

= - \frac{1}{6} ln tan (\frac{1}{2} arctan (\frac{1}{3} x)) + C

作图

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