integral of (2x)/((x-1)(x+3))

解答

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

2 (\frac{1}{4} ln ∣ x - 1 ∣ + \frac{3}{4} ln ∣ x + 3 ∣) + C

求解步骤

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

提出常数:

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

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

将

\frac{x}{( x - 1 ) ( x + 3 )}

用部份分式展开:

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

= 2 \cdot \int \frac{1}{4 ( x - 1 )} + \frac{3}{4 ( x + 3 )} 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}{4 ( x - 1 )} d x + \int \frac{3}{4 ( x + 3 )} d x)

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

\int \frac{3}{4 ( x + 3 )} d x = \frac{3}{4} ln ∣ x + 3 ∣

= 2 (\frac{1}{4} ln ∣ x - 1 ∣ + \frac{3}{4} ln ∣ x + 3 ∣)

解答补常数

= 2 (\frac{1}{4} ln ∣ x - 1 ∣ + \frac{3}{4} ln ∣ x + 3 ∣) + C

d er i v a t i v e x^{6} e^{x}

t an g e n t f (x) = 4 x^{2} + 2 x, (- 3, 30)

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

d er i v a t i v e (x - 3) 4 - x

\frac{d}{d x} ((1 - x^{3})^{\frac{1}{2}})