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

解答

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

- \frac{1}{12} ln ∣ x + 1 ∣ - \frac{5}{3} ln ∣ x - 2 ∣ + \frac{7}{4} ln ∣ x - 3 ∣ + C

求解步骤

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

将

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

用部份分式展开:

- \frac{1}{12 ( x + 1 )} - \frac{5}{3 ( x - 2 )} + \frac{7}{4 ( x - 3 )}

= \int - \frac{1}{12 ( x + 1 )} - \frac{5}{3 ( x - 2 )} + \frac{7}{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

= - \int \frac{1}{12 ( x + 1 )} d x - \int \frac{5}{3 ( x - 2 )} d x + \int \frac{7}{4 ( x - 3 )} d x

\int \frac{1}{12 ( x + 1 )} d x = \frac{1}{12} ln ∣ x + 1 ∣

\int \frac{5}{3 ( x - 2 )} d x = \frac{5}{3} ln ∣ x - 2 ∣

\int \frac{7}{4 ( x - 3 )} d x = \frac{7}{4} ln ∣ x - 3 ∣

= - \frac{1}{12} ln ∣ x + 1 ∣ - \frac{5}{3} ln ∣ x - 2 ∣ + \frac{7}{4} ln ∣ x - 3 ∣

解答补常数

= - \frac{1}{12} ln ∣ x + 1 ∣ - \frac{5}{3} ln ∣ x - 2 ∣ + \frac{7}{4} ln ∣ x - 3 ∣ + C

\int 7 - 2 x d x

\frac{d}{d x} (\frac{3 x ^{2} tan ( x )}{sec ( x )})

d er i v a t i v e sin (cos (tan (π x)))

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

x \to 2 + lim (\frac{x}{x ^{2} + 4})