integral of 2/((x-4)(x^2+2x+6))

解答

\int \frac{2}{( x - 4 ) ( x ^{2} + 2 x + 6 )} d x

2 (\frac{1}{30} ln ∣ x - 4 ∣ - \frac{1}{60} ln x^{2} + 2 x + 6 - \frac{1}{6 5} arctan (\frac{1}{5} (x + 1))) + C

求解步骤

\int \frac{2}{( x - 4 ) ( x ^{2} + 2 x + 6 )} d x

提出常数:

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

= 2 \cdot \int \frac{1}{( x - 4 ) ( x ^{2} + 2 x + 6 )} d x

将

\frac{1}{( x - 4 ) ( x ^{2} + 2 x + 6 )}

用部份分式展开:

\frac{1}{30 ( x - 4 )} + \frac{- x - 6}{30 ( x ^{2} + 2 x + 6 )}

= 2 \cdot \int \frac{1}{30 ( x - 4 )} + \frac{- x - 6}{30 ( x ^{2} + 2 x + 6 )} d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{1}{30 ( x - 4 )} d x + \int \frac{- x - 6}{30 ( x ^{2} + 2 x + 6 )} d x)

\int \frac{1}{30 ( x - 4 )} d x = \frac{1}{30} ln ∣ x - 4 ∣

\int \frac{- x - 6}{30 ( x ^{2} + 2 x + 6 )} d x = \frac{1}{60 5} (- 5 ln x^{2} + 2 x + 6 - 10 arctan (\frac{1}{5} (x + 1)))

= 2 (\frac{1}{30} ln ∣ x - 4 ∣ + \frac{1}{60 5} (- 5 ln x^{2} + 2 x + 6 - 10 arctan (\frac{1}{5} (x + 1))))

乘开

\frac{1}{60 5} (- 5 ln x^{2} + 2 x + 6 - 10 arctan (\frac{1}{5} (x + 1))) : - \frac{1}{60} ln x^{2} + 2 x + 6 - \frac{1}{6 5} arctan (\frac{1}{5} (x + 1))

= 2 (\frac{1}{30} ln ∣ x - 4 ∣ - \frac{1}{60} ln x^{2} + 2 x + 6 - \frac{1}{6 5} arctan (\frac{1}{5} (x + 1)))

解答补常数

= 2 (\frac{1}{30} ln ∣ x - 4 ∣ - \frac{1}{60} ln x^{2} + 2 x + 6 - \frac{1}{6 5} arctan (\frac{1}{5} (x + 1))) + C