integral of (2x-7)/((x+1)(x-3))

解答

\int \frac{2 x - 7}{( x + 1 ) ( x - 3 )} d x

\frac{9}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 3 ∣ + C

求解步骤

\int \frac{2 x - 7}{( x + 1 ) ( x - 3 )} d x

乘开

\frac{2 x - 7}{( x + 1 ) ( x - 3 )} : \frac{2 x}{( x + 1 ) ( x - 3 )} - \frac{7}{( x + 1 ) ( x - 3 )}

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{2 x}{( x + 1 ) ( x - 3 )} d x - \int \frac{7}{( x + 1 ) ( x - 3 )} d x

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

\int \frac{7}{( x + 1 ) ( x - 3 )} d x = 7 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣)

= 2 (\frac{1}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 3 ∣) - 7 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣)

化简

2 (\frac{1}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 3 ∣) - 7 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣) : \frac{9}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 3 ∣

= \frac{9}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 3 ∣

解答补常数

= \frac{9}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 3 ∣ + C

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

\int \frac{4}{x x} d x

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

\int \frac{x}{3 x ^{2} + 7} d x

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