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

解答

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

- \frac{5}{4} ln ∣ x - 1 ∣ + \frac{5}{4} ln ∣ x + 1 ∣ - \frac{1}{x - 1} - \frac{7}{2 ( x - 1 )} + C

求解步骤

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

乘开

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

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

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

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

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

化简

\frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{x - 1} - \frac{1}{2} ln ∣ x + 1 ∣ + 7 (- \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} + \frac{1}{4} ln ∣ x + 1 ∣) : - \frac{5}{4} ln ∣ x - 1 ∣ + \frac{5}{4} ln ∣ x + 1 ∣ - \frac{1}{x - 1} - \frac{7}{2 ( x - 1 )}

= - \frac{5}{4} ln ∣ x - 1 ∣ + \frac{5}{4} ln ∣ x + 1 ∣ - \frac{1}{x - 1} - \frac{7}{2 ( x - 1 )}

解答补常数

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

\int_{1}^{2} 18 x^{2} 4^{x^{3}} d x

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

x \to 0 - lim (1 + e^{\frac{1}{x}})

l a pl a ce t r an s f or m 3 t^{2} + 15 t + 15

s l o p e y = (5 x^{2} - 1)^{- 3}