integral of ln(2x-1)

解答

\int ln (2 x - 1) d x

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

求解步骤

\int ln (2 x - 1) d x

使用换元积分法

= \int ln (u) \frac{1}{2} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int ln (u) d u

使用分布积分法

= \frac{1}{2} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{2} (u ln (u) - u)

u = 2 x - 1

代回

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

化简

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

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

解答补常数

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

\frac{d}{d x} (2 x^{3} - 3 x)

\frac{d}{d x} (arcsin (\frac{y}{x ^{2} + y ^{2}}))

\int \frac{3 x - 4}{x ^{2} - 2 x} d x

\int 4 x^{3} - 3 x^{2} + 5 d x

\int (3 x^{2} y^{- 2} + 2 x y^{- 1}) d x