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受欢迎的微积分 >

integral of 1/(x^2-6x+25)

解答

\int \frac{1}{x ^{2} - 6 x + 25} d x

解答

\frac{1}{4} arctan (\frac{x - 3}{4}) + C

求解步骤

\int \frac{1}{x ^{2} - 6 x + 25} d x

对

x^{2} - 6 x + 25

配方:

(x - 3)^{2} + 16

= \int \frac{1}{( x - 3 ) ^{2} + 16} d x

使用换元积分法

= \int \frac{1}{u ^{2} + 16} d u

使用换元积分法

= \int \frac{1}{4 ( 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}{4} \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}{4} arctan (v)

代回

= \frac{1}{4} arctan (\frac{x - 3}{4})

解答补常数

= \frac{1}{4} arctan (\frac{x - 3}{4}) + C

作图

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