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

解答

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

2 (\frac{1}{20} ln ∣ x - 2 ∣ + \frac{1}{40} (- ln x^{2} + 2 x + 2 - arctan (x + 1)) - \frac{1}{8} arctan (x - 1)) + C

求解步骤

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

将

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

用部份分式展开:

\frac{1}{20 ( x - 2 )} + \frac{- 2 x - 3}{40 ( x ^{2} + 2 x + 2 )} - \frac{1}{8 ( x ^{2} - 2 x + 2 )}

= 2 \cdot \int \frac{1}{20 ( x - 2 )} + \frac{- 2 x - 3}{40 ( x ^{2} + 2 x + 2 )} - \frac{1}{8 ( x ^{2} - 2 x + 2 )} 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}{20 ( x - 2 )} d x + \int \frac{- 2 x - 3}{40 ( x ^{2} + 2 x + 2 )} d x - \int \frac{1}{8 ( x ^{2} - 2 x + 2 )} d x)

\int \frac{1}{20 ( x - 2 )} d x = \frac{1}{20} ln ∣ x - 2 ∣

\int \frac{- 2 x - 3}{40 ( x ^{2} + 2 x + 2 )} d x = \frac{1}{40} (- ln x^{2} + 2 x + 2 - arctan (x + 1))

\int \frac{1}{8 ( x ^{2} - 2 x + 2 )} d x = \frac{1}{8} arctan (x - 1)

= 2 (\frac{1}{20} ln ∣ x - 2 ∣ + \frac{1}{40} (- ln x^{2} + 2 x + 2 - arctan (x + 1)) - \frac{1}{8} arctan (x - 1))

解答补常数

= 2 (\frac{1}{20} ln ∣ x - 2 ∣ + \frac{1}{40} (- ln x^{2} + 2 x + 2 - arctan (x + 1)) - \frac{1}{8} arctan (x - 1)) + C

x \to - \infty lim (- 3 - x)

\int_{0}^{3} x^{2} + 16 d x

\frac{d y}{d x} + \frac{y}{x} = \frac{- y ^{2}}{x}

\int \frac{4 x}{( 2 x ^{2} + 3 ) ^{3}} d x

\int \frac{x ^{2} - 3}{x + 1} d x