integral of (ln(x)-1)/x

解答

\int \frac{ln ( x ) - 1}{x} d x

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

求解步骤

\int \frac{ln ( x ) - 1}{x} d x

使用换元积分法

= \int u d u

使用幂法则

= \frac{u ^{2}}{2}

u = ln (x) - 1

代回

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

解答补常数

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

x \to 1 lim (3 x - 6)

\frac{d}{d x} (ln (6 \frac{x - 5}{x + 5}))

in v erse l a pl a ce \frac{2 s e ^{- 2 s}}{5 s ^{2} + 45}

\frac{d y}{d x} = \frac{y}{( 1 + x ^{2} )}

y^{^{''}} - y^{^{'}} - 2 y = 4 e^{- t}