integral of (3x^2+1)/(x^5-x)

解答

\int \frac{3 x ^{2} + 1}{x ^{5} - x} d x

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

求解步骤

\int \frac{3 x ^{2} + 1}{x ^{5} - x} d x

乘开

\frac{3 x ^{2} + 1}{x ^{5} - x} : \frac{3 x ^{2}}{x ^{5} - x} + \frac{1}{x ^{5} - 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{3 x ^{2}}{x ^{5} - x} d x + \int \frac{1}{x ^{5} - x} d x

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

\int \frac{1}{x ^{5} - x} d x = - ln ∣ x ∣ + \frac{1}{4} ln x^{2} + 1 + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣

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

化简

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

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

解答补常数

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

s l o p e in t erce pt (0.5) (- 2.4)

y^{^{'}} = 5 y + y^{6}, y (0) = 1

\int_{0}^{x} \frac{1}{t ^{2} + t + 1} d t

x \to 0 lim (sec (2 x))

in v erse l a pl a ce \frac{1}{s - 4}