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

解答

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

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

求解步骤

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

将

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

用部份分式展开:

- \frac{2}{x} - \frac{1}{x ^{2}} + \frac{4}{2 x - 1}

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

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

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

化简

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

解答补常数

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

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

d er i v a t i v e 2 e^{(x - 5)}

d er i v a t i v e \frac{x - 4}{x + 4}

x \to 0 lim (sin (\frac{40 π}{x}))

in v erse l a pl a ce \frac{1}{( s + 2 ) ^{4}}