integral of ln|x-1|

解答

\int ln ∣ x - 1 ∣ d x

x ln ∣ x - 1 ∣ - x + 1 - ln ∣ x - 1 ∣ + C

求解步骤

\int ln ∣ x - 1 ∣ d x

使用分布积分法

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

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

= x ln ∣ x - 1 ∣ - (x - 1 + ln ∣ x - 1 ∣)

化简

= x ln ∣ x - 1 ∣ - x + 1 - ln ∣ x - 1 ∣

解答补常数

= x ln ∣ x - 1 ∣ - x + 1 - ln ∣ x - 1 ∣ + C

d er i v a t i v e f (x) = \frac{x}{1 - x ^{2}}

d er i v a t i v e f (x) = \frac{2500 x}{ln ( 8 x + 8 )}

x y^{^{'}} + (x + \frac{1}{3}) y = e^{- 2 x} y^{- 2}

t an g e n t f (x) = x^{2} - x + 4, at x = 1

ma c l a u r in \frac{1}{3 x + 8}