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受欢迎的微积分 >

integral of 1/(9x^2-18x+72)

解答

\int \frac{1}{9 x ^{2} - 18 x + 72} d x

解答

\frac{1}{9 7} arctan (\frac{x - 1}{7}) + C

求解步骤

\int \frac{1}{9 x ^{2} - 18 x + 72} d x

对

9 x^{2} - 18 x + 72

配方:

9 (x - 1)^{2} + 63

= \int \frac{1}{9 ( x - 1 ) ^{2} + 63} d x

使用换元积分法

= \int \frac{1}{9 u ^{2} + 63} d u

使用换元积分法

= \int \frac{1}{9 7 ( v ^{2} + 1 )} d v

提出常数:

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

= \frac{1}{9 7} \cdot \int \frac{1}{v ^{2} + 1} d v

使用积分常见定则:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{9 7} arctan (v)

代回

= \frac{1}{9 7} arctan (\frac{3}{3 7} (x - 1))

化简

\frac{1}{9 7} arctan (\frac{3}{3 7} (x - 1)) : \frac{1}{9 7} arctan (\frac{x - 1}{7})

= \frac{1}{9 7} arctan (\frac{x - 1}{7})

解答补常数

= \frac{1}{9 7} arctan (\frac{x - 1}{7}) + C

作图

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