integral of 1/(x^3+9x)

解答

\int \frac{1}{x ^{3} + 9 x} d x

\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9 + C

求解步骤

\int \frac{1}{x ^{3} + 9 x} d x

将

\frac{1}{x ^{3} + 9 x}

用部份分式展开:

\frac{1}{9 x} - \frac{x}{9 ( x ^{2} + 9 )}

= \int \frac{1}{9 x} - \frac{x}{9 ( x ^{2} + 9 )} 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}{9 x} d x - \int \frac{x}{9 ( x ^{2} + 9 )} d x

\int \frac{1}{9 x} d x = \frac{1}{9} ln ∣ x ∣

\int \frac{x}{9 ( x ^{2} + 9 )} d x = \frac{1}{18} ln x^{2} + 9

= \frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9

解答补常数

= \frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9 + C

t an g e n t 5 sec (x), at x = \frac{π}{3}

a re a - x^{2} + 3 x, 2 x^{3} - x^{2} - 5 x

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

\int \frac{y}{( y ^{2} + z ^{2} ) ^{\frac{3}{2}}} d y

s l o p e (1, 7), (7, 3)